# HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

#### blyatman

##### Well-Known Member
These are all beyond the scope of the syllabus.
Yes, please see one of the subsequent posts.

#### stupid_girl

##### Active Member
I have a feeling this is not the fastest method...

$\bg_white I = \int_0^1\left( \arcsin\left(\frac{x}{x+1}\right)\right)^2\,dx$

$\bg_white =\int_0^{\pi/6} \frac{y^2\cos y}{(1-\sin y)^2}\,dy \quad \left(y = \arcsin \left(\frac{x}{x+1}\right) \iff x = \frac{\sin y}{1-\sin y }\right)$

$\bg_white =\frac{\pi^2}{18}- 2 \int_0^{\pi/6} \frac{y}{1-\sin y}\,dy \quad \left({\rm IBP:}\,u = y^2, \, dv = \frac{\cos y}{(1-\sin y)^2}\, dy\right)$

$\bg_white \text{Now,}\, \int \frac{1}{1-\sin y}\,dy = \frac{1}{\tan(y/2)-1} + C.$

(the proof is left as an exercise to the reader.)

$\bg_white \implies I =\frac{\pi^2}{18} - 2\left(-\frac{\pi/3}{(\tan(\pi/12) - 1)} +2 \int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy\right) \quad \left({\rm IBP:}\,u = y, \, dv = \frac{1}{1-\sin y}\, dy\right).$

From here, I will use the result $\bg_white \tan \pi/12 = 2 - \sqrt 3$, which is not difficult to verify via the appropriate double angle identity.

$\bg_white \int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy$

$\bg_white = \int_0^{2 - \sqrt 3} \frac{2}{(t-1)(1+t^2)}\, dt \quad \left(t = \tan \frac y2\right)$

$\bg_white = \int_0^{2 - \sqrt 3} \, -\frac{t}{t^2+1}- \frac{1}{t^2+1} + \frac{1}{t-1}\, dt$

$\bg_white = \log\left( \frac{{|t-1|} }{\sqrt{t^2+1}} \right) - \arctan t \Bigg|_0^{2 - \sqrt 3}$

$\bg_white =-\frac 12 \log 2 - \frac{\pi}{12}.$

Note that the absolute value is important when evaluating the lower bound.

$\bg_white \implies I = \frac{\pi^2}{18} - 2\left(\frac \pi 6 + \frac{\pi \sqrt 3}{6}\right)+ 4 \left( \frac 12 \log 2 + \frac \pi {12}\right)$

$\bg_white = \frac{\pi^2}{18} - \frac{\pi \sqrt 3}{3}+ 2 \log 2$

$\bg_white \approx 0.1208.$
My method is similar to yours. DI table may make the integration by parts neater.
$\bg_white D \qquad I$
$\bg_white x^2 \qquad \frac{\cos x}{\left(1-\sin x\right)^2}$
$\bg_white 2x\qquad \frac{1}{1-\sin x}$
$\bg_white 2\qquad \frac{\cos x}{1-\sin x}$
$\bg_white 0\qquad -\ln\left(1-\sin x\right)$

$\bg_white \int_{ }^{ }\frac{x^2\cos x}{\left(1-\sin x\right)^2}dx=\frac{x^2}{1-\sin x}-\frac{2x\cos x}{1-\sin x}-2\ln\left(1-\sin x\right)+C$

#### stupid_girl

##### Active Member
This one is challenging if you can't think of a useful substitution.
$\bg_white \int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx$
Have anyone figured out?

The trick is:
$\bg_white \int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx$
$\bg_white =\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{8\left(2+\sqrt{5}\right)}{27}\right)^x+\left(\frac{6\left(3+\sqrt{5}\right)}{27}\right)^x}dx$
$\bg_white =\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{1+\sqrt{5}}{3}\right)^{3x}+\left(\frac{1+\sqrt{5}}{3}\right)^{2x}}dx$

Clearly, the useful substitution is
$\bg_white u=\left(\frac{1+\sqrt{5}}{3}\right)^x$

#### stupid_girl

##### Active Member
This one is relatively routine.
$\bg_white \int_0^{\frac{\pi}{2}}\frac{x\cos x+6x\sin x}{\left(5\sin x-30\cos x+31\right)^2}dx=\frac{\pi}{72}$
This integral itself should be quite routine. Getting the final answer in terms of pi is slightly tricky.

$\bg_white \int\frac{x\cos x+6x\sin x}{\left(5\sin x-30\cos x+31\right)^2}dx$
$\bg_white =\frac{1}{5}\int x\frac{\left(5\cos x+30\sin x\right)}{\left(5\sin x-30\cos x+31\right)^2}dx$
$\bg_white =-\frac{1}{5}\int x\ d\left(\frac{1}{5\sin x-30\cos x+31}\right)$
$\bg_white =-\frac{x}{5}\left(\frac{1}{5\sin x-30\cos x+31}\right)+\frac{1}{5}\int_{ }^{ }\left(\frac{1}{5\sin x-30\cos x+31}\right)dx$
$\bg_white =-\frac{x}{5}\left(\frac{1}{5\sin x-30\cos x+31}\right)+\frac{1}{15}\tan^{-1}\left(\frac{61\tan\frac{x}{2}+5}{6}\right)+c$

To get the final answer, you need to show that:
$\bg_white \tan^{-1}11-\tan^{-1}\frac{5}{6}=\frac{\pi}{4}$

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##### -insert title here-
Clearly, the useful substitution is
$\bg_white u=\left(\frac{1+\sqrt{5}}{3}\right)^x$
intuition said it was φ-like but i didn't want to stay up that late to figure it out;

a good one nonetheless

#### blyatman

##### Well-Known Member
$\bg_white 1.~~\int_{0}^{\infty}\frac{dx}{(1+x^2)^2}~~~(\mathrm{Ans}:~\frac{\pi}{4})$
$\bg_white 2.~~\int_{0}^{\infty}\frac{x}{(1+x^2)^2}dx~~~(\mathrm{Ans}:~\frac{1}{2})$
$\bg_white 3.~~\int_{0}^{\infty}\frac{dx}{1+x^4}~~~(\mathrm{Ans}:~\frac{\pi\sqrt{2}}{4})$
$\bg_white 4.~~\int_{0}^{\infty}\frac{x^2}{x^6+1}dx~~~(\mathrm{Ans}:~\frac{\pi}{6})$
$\bg_white 5.~~\int_{0}^{\infty}\frac{x^{1/3}}{1+x^2}dx~~~(\mathrm{Ans}:~\frac{\pi}{\sqrt{6}})$
$\bg_white 6.~~\int_{0}^{2\pi}\frac{\cos^2\theta}{5-4\cos\theta}d\theta~~~(\mathrm{Ans}:~\frac{5\pi}{12})$
$\bg_white 7.~~\int_{0}^{2\pi}\frac{\cos3\theta}{5-4\cos\theta}d\theta~~~(\mathrm{Ans}:~\frac{\pi}{12})$
$\bg_white 8.~~\int_{0}^{\infty}\frac{\cos sx}{k^2+x^2}dx~~~(\mathrm{Ans}:~\frac{\pi}{2k}e^{-ks})$
$\bg_white 9.~~\int_{0}^{\infty}\frac{\cos x}{x^2+a^2}dx~~~(\mathrm{Ans}:~\frac{\pi e^{-a}}{2a})$
$\bg_white 10.~~\int_{0}^{\infty}\frac{\cos 2x}{(x^2+1)^2}dx$
$\bg_white 11.~~\int_{0}^{\infty}\frac{cos 4x}{x^4+5x^2+4}dx$
$\bg_white 12.~~\int_{0}^{\infty}\frac{x\sin x}{x^2+4}dx~~~(\mathrm{Ans}:~\frac{\pi}{4e}\sin1)$

These are Mr Blyatman's question. Ill work them up later today
As it turns out, the indefinite counterparts of some of these integrals have no closed-form solutions (I tried googling a solution to no.9), so some of these definite integrals must be solved (assuming they can be solved conventionally) without directly integrating into the anti-derivative, similar to how the integral of exp(x)cos(x) is done by using integration by parts twice (so that we have 2*integral = something).

#### HeroWise

##### Active Member
Have anyone figured out?

The trick is:
$\bg_white \int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx$
$\bg_white =\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{8\left(2+\sqrt{5}\right)}{27}\right)^x+\left(\frac{6\left(3+\sqrt{5}\right)}{27}\right)^x}dx$
$\bg_white =\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{1+\sqrt{5}}{3}\right)^{3x}+\left(\frac{1+\sqrt{5}}{3}\right)^{2x}}dx$

Clearly, the useful substitution is
$\bg_white u=\left(\frac{1+\sqrt{5}}{3}\right)^x$
How are you supposed to know phi is a good sub here? I mean looking at the original question alone

#### DrEuler

##### Member
How are you supposed to know phi is a good sub here? I mean looking at the original question alone
Inspection.

#### stupid_girl

##### Active Member
How are you supposed to know phi is a good sub here? I mean looking at the original question alone
After dividing top and bottom by 27^x, the integral is in the form 1/(p^x+q^x+r^x). If this integral has an elementary form, then there must be a relationship between p,q and r. After some trial and error, you would get that relationship.

#### stupid_girl

##### Active Member
This one is tedious.
$\bg_white \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{x^2-2\pi+\pi^2}{\left(\pi\sin x+x\right)^2+\left(\pi+x\right)\left(\pi-x\right)\cos x+\pi^2\cos^2x}dx=\frac{1037\sqrt{2}+259\sqrt{3}-1893}{427}$
Has anyone attempted this?
$\bg_white \int\frac{x^2-2\pi+\pi^2}{\left(\pi\sin x+x\right)^2+\left(\pi+x\right)\left(\pi-x\right)\cos x+\pi^2\cos^2x}dx$
$\bg_white =\int\frac{x^2-2\pi+\pi^2}{2\left(x\sin\frac{x}{2}+\pi\cos\frac{x}{2}\right)^2}dx$
$\bg_white =\int\frac{x^2-2\pi+\pi^2}{2\left(\sqrt{x^2+\pi^2}\cos\left(\frac{x}{2}-\tan^{-1}\frac{x}{\pi}\right)\right)^2}dx$
$\bg_white =\int\left(\frac{1}{2}-\frac{\pi}{x^2+\pi^2}\right)\sec^2\left(\frac{x}{2}-\tan^{-1}\frac{x}{\pi}\right)dx$
$\bg_white =\tan\left(\frac{x}{2}-\tan^{-1}\frac{x}{\pi}\right)+c$
$\bg_white =\frac{\tan\frac{x}{2}-\frac{x}{\pi}}{1+\frac{x}{\pi}\tan\frac{x}{2}}+c$

To get the final answer, you have to express the annoying tangent in surd form.
$\bg_white \tan\frac{\pi}{12}=2-\sqrt{3}$
$\bg_white \tan\frac{\pi}{8}=\sqrt{2}-1$

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#### stupid_girl

##### Active Member
Feel free to share your attempt.

$\bg_white \int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$\bg_white =\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$

##### -insert title here-
If this integral has an elementary form, then there must be a relationship between p,q and r. After some trial and error, you would get that relationship.
That... doesn't necessarily have to be true, but my intuition agrees with that.

Formally, however, I do not.

#### vernburn

##### New Member

Just responding to an earlier integral based on the infinite expansion for e. I just discovered this thread and i love it!

MIT ???

#### stupid_girl

##### Active Member
Feel free to share your attempt.

$\bg_white \int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$\bg_white =\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$
No attempt for 2 weeks
Is it too tedious? I think it's a good practice for algebraic skills.

#### Drdusk

##### π
Moderator
Feel free to share your attempt.

$\bg_white \int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$\bg_white =\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$
Math is not really my area but I tried nonetheless

$\bg_white \int_{0.25}^{0.5} log_2(ex)sin(x^x + ln\sqrt{x})cos(x^x - ln\sqrt{x})dx$

$\bg_white = \frac{1}{2}\int_{0.25}^{0.5}log_2(ex)\bigg(sin(2x^x) + sin(2ln\sqrt{x})\bigg)dx$

$\bg_white = \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(2x^x)dx + \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$\bg_white I = \frac{1}{2ln(2)}\int (1+lnx)sin(2x^x)dx$

$\bg_white u = 2x^x \Rightarrow \frac{du}{u} = 2(1 + lnx)dx$

$\bg_white \therefore I = \frac{1}{4ln(2)}\int usin(u)du = \frac{1}{4ln(2)} \bigg[sin(2x^x) - 2x^xcos(2x^x)\bigg]_{0.25}^{0.5} = 0$

$\bg_white K = \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$\bg_white u = ln(x)\Rightarrow du = \frac{1}{x}dx$

$\bg_white \therefore K = \frac{1}{2ln(2)}\int (1 + u)e^usin(u)$

$\bg_white = \frac{1}{2ln(2)}\int e^usin(u)du + \frac{1}{2ln(2)}\int ue^usin(u)du$

$\bg_white J = \frac{1}{2ln(2)}\int e^xsin(x)dx\hspace{2mm}\text{Replacing u with x for simplicity}$

$\bg_white J = \frac{1}{2ln(2)}\bigg(e^xsin(x) - \int e^xcos(x)\bigg)$

$\bg_white J= \frac{1}{2ln(2)}\bigg(e^xsin(x) - e^xcos(x) - 2ln(2)J\bigg)$

$\bg_white J = \frac{1}{4ln(2)}\Bigg[x\bigg(sin\big(lnx\big)-cos\big(lnx\big)\bigg)\Bigg]_{0.25}^{0.5}$

$\bg_white = \frac{1}{4ln(2)}\Bigg(\frac{1}{2}sin(-ln2) - \frac{1}{2}cos(-ln2) - \frac{1}{4}sin(-ln4) + \frac{1}{4}cos(-ln4)\Bigg)$

$\bg_white \text{Now for the second integral}\hspace{2mm}S = \frac{1}{2ln(2)}\int ue^usin(u)du = \frac{1}{2ln(2)}\int xe^xsin(x)dx\hspace{2mm}\text{For simplicity}$

$\bg_white u = xe^x\Rightarrow du = (e^x + xe^x)dx, \hspace{2mm}dv = sin(x)dx\Rightarrow v = -cos(x)$

$\bg_white \therefore S = -xe^xcos(x) + \int e^xcos(x)dx + \int xe^xcos(x)dx$

$\bg_white S = -S + \frac{e^x(cos(x) - sin(x))}{4ln(2)} - \frac{xe^xcos(x)}{2ln(2)} -\frac{xe^xsin(x)}{2ln(2)} - \frac{1}{2ln(2)}\int e^xsin(x)dx$

$\bg_white \therefore S = \frac{e^x(cos(x) - sin(x))}{8ln(2)} - \frac{xe^xcos(x)}{4ln(2)} -\frac{xe^xsin(x)}{4ln(2)} - \frac{1}{4ln(2)}\int e^xsin(x)dx$

$\bg_white \text{But we replaced x with u before for simplicity, so we must now convert it all back which gives us }$

$\bg_white S = \bigg[\frac{x\big(cos(lnx) - sin(lnx)\big)}{8ln(2)}\bigg]_{0.25}^{0.5} - \bigg[\frac{xlnx(cos(lnx) + sin(lnx))}{4ln(2)}\bigg]_{0.25}^{0.5}-\bigg[\frac{x(sin(lnx)-cos(lnx))}{8ln(2)}\bigg]_{0.25}^{0.5}$

$\bg_white \therefore S = \frac{1}{8ln(2)}\bigg[\frac{cos(-ln2)- sin(-ln2)}{2} - \frac{cos(-ln4) - sin(ln4)}{4}\bigg] - \frac{1}{4ln(2)}\bigg[\frac{-ln(2)(cos(-ln2) + sin(-ln2))}{2} + \frac{ln(4)(cos(-ln4) + sin(-ln4))}{4}\bigg] - \frac{1}{8ln(2)}\bigg[\frac{sin(-ln2)- cos(-ln2)}{2} - \frac{sin(-ln4) - cos(ln4)}{4}\bigg]$

$\bg_white \text{Rest is trivial algebraic elimination/addition which I can't type rn ;-;. Thank you for ruining my brain and hand}$

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#### HeroWise

##### Active Member
Math is not really my area but I tried nonetheless

$\bg_white \int_{0.25}^{0.5} log_2(ex)sin(x^x + ln\sqrt{x})cos(x^x - ln\sqrt{x})dx$

$\bg_white = \frac{1}{2}\int_{0.25}^{0.5}log_2(ex)\bigg(sin(2x^x) + sin(2ln\sqrt{x})\bigg)dx$

$\bg_white = \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(2x^x)dx + \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$\bg_white I = \frac{1}{2ln(2)}\int (1+lnx)sin(2x^x)dx$

$\bg_white u = 2x^x \Rightarrow \frac{du}{u} = 2(1 + lnx)dx$

$\bg_white \therefore I = \frac{1}{4ln(2)}\int usin(u)du = \frac{1}{4ln(2)} \bigg[sin(2x^x) - 2x^xcos(2x^x)\bigg]_{0.25}^{0.5} = 0$

$\bg_white K = \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$\bg_white u = ln(x)\Rightarrow du = \frac{1}{x}dx$

$\bg_white \therefore K = \frac{1}{2ln(2)}\int (1 + u)e^usin(u)$

$\bg_white = \frac{1}{2ln(2)}\int e^usin(u)du + \frac{1}{2ln(2)}\int ue^usin(u)du$

$\bg_white J = \frac{1}{2ln(2)}\int e^xsin(x)dx\hspace{2mm}\text{Replacing u with x for simplicity}$

$\bg_white J = \frac{1}{2ln(2)}\bigg(e^xsin(x) - \int e^xcos(x)\bigg)$

$\bg_white J= \frac{1}{2ln(2)}\bigg(e^xsin(x) - e^xcos(x) - 2ln(2)J\bigg)$

$\bg_white J = \frac{1}{4ln(2)}\Bigg[x\bigg(sin\big(lnx\big)-cos\big(lnx\big)\bigg)\Bigg]_{0.25}^{0.5}$

$\bg_white = \frac{1}{4ln(2)}\Bigg(\frac{1}{2}sin(-ln2) - \frac{1}{2}cos(-ln2) - \frac{1}{4}sin(-ln4) + \frac{1}{4}cos(-ln4)\Bigg)$

$\bg_white \text{Now for the second integral}\hspace{2mm}S = \frac{1}{2ln(2)}\int ue^usin(u)du = \frac{1}{2ln(2)}\int xe^xsin(x)dx\hspace{2mm}\text{For simplicity}$

$\bg_white u = xe^x\Rightarrow du = (e^x + xe^x)dx, \hspace{2mm}dv = sin(x)dx\Rightarrow v = -cos(x)$

$\bg_white \therefore S = -xe^xcos(x) + \int e^xcos(x)dx + \int xe^xcos(x)dx$

$\bg_white S = -S + \frac{e^x(cos(x) - sin(x))}{4ln(2)} - \frac{xe^xcos(x)}{2ln(2)} -\frac{xe^xsin(x)}{2ln(2)} - \frac{1}{2ln(2)}\int e^xsin(x)dx$

$\bg_white \therefore S = \frac{e^x(cos(x) - sin(x))}{8ln(2)} - \frac{xe^xcos(x)}{4ln(2)} -\frac{xe^xsin(x)}{4ln(2)} - \frac{1}{4ln(2)}\int e^xsin(x)dx$

$\bg_white \text{But we replaced x with u before for simplicity, so we must now convert it all back which gives us }$

$\bg_white S = \bigg[\frac{x\big(cos(lnx) - sin(lnx)\big)}{8ln(2)}\bigg]_{0.25}^{0.5} - \bigg[\frac{xlnx(cos(lnx) + sin(lnx))}{4ln(2)}\bigg]_{0.25}^{0.5}-\bigg[\frac{x(sin(lnx)-cos(lnx))}{8ln(2)}\bigg]_{0.25}^{0.5}$

$\bg_white \therefore S = \frac{1}{8ln(2)}\bigg[\frac{cos(-ln2)- sin(-ln2)}{2} - \frac{cos(-ln4) - sin(ln4)}{4}\bigg] - \frac{1}{4ln(2)}\bigg[\frac{-ln(2)(cos(-ln2) + sin(-ln2))}{2} + \frac{ln(4)(cos(-ln4) + sin(-ln4))}{4}\bigg] - \frac{1}{8ln(2)}\bigg[\frac{sin(-ln2)- cos(-ln2)}{2} - \frac{sin(-ln4) - cos(ln4)}{4}\bigg]$

$\bg_white \text{Rest is trivial algebraic elimination/addition which I can't type rn ;-;. Thank you for ruining my brain and hand}$

"Math is not my area",,,

Bro that question legit took an hour to solve

#### stupid_girl

##### Active Member
Good try!
Unfortunately, there's a mistake in this piece of integral.
Fortunately, it doesn't affect the final answer.
$\bg_white I = \frac{1}{2ln(2)}\int (1+lnx)sin(2x^x)dx$

$\bg_white u = 2x^x \Rightarrow \frac{du}{u} = 2(1 + lnx)dx$

$\bg_white \therefore I = \frac{1}{4ln(2)}\int usin(u)du = \frac{1}{4ln(2)} \bigg[sin(2x^x) - 2x^xcos(2x^x)\bigg]_{0.25}^{0.5} = 0$

#### Drdusk

##### π
Moderator
Good try!
Unfortunately, there's a mistake in this piece of integral.
Fortunately, it doesn't affect the final answer.
yeah I kinda realized that while going to sleep but I have no idea how to integrate sinx/x. so I kinda just went meh

#### stupid_girl

##### Active Member
It should be well known enough that
$\bg_white \int e^x\sin x=\frac{e^x}{2}\left(\sin\ x-\cos x\right)+c$
and
$\bg_white \int e^x\cos x=\frac{e^x}{2}\left(\sin\ x+\cos x\right)+c$
(can be derived by integration by parts twice)

This is how you may make the life easier.
$\bg_white \int_{ }^{ }\left(x+1\right)e^x\sin xdx$
$\bg_white =\frac{1}{2}\int\left(x+1\right)d\left(e^x\left(\sin\ x-\cos x\right)\right)$
$\bg_white =\frac{1}{2}\left(x+1\right)\left(e^x\left(\sin\ x-\cos x\right)\right)-\frac{1}{2}\int\left(e^x\left(\sin\ x-\cos x\right)\right)dx$

Overall speaking, the integral requires log rule, trig rule, 2 substitutions, 3 times integration by parts and algebraic manipulations...nice practice for MX2.