# Urgent questions help (1 Viewer)

#### Farhanthestudent005

##### Member
Can someone send me solutions for these 3 questions? Will appreciate it.

#### cossine

##### Active Member
Use an online integral/differentiation calculator

#### 5uckerberg

##### Well-Known Member
$\bg_white \int_{0}^{\frac{\pi}{4}}\frac{\sin{2x}}{4+\cos^{2}{x}dx$
Let $\bg_white u=\cos^{2}{x}, du=-2\sin{x}\cos{x}$
$\bg_white du=-\sin{2x}dx$
When $\bg_white x=\frac{\pi}{4}, u=\frac{1}{2} , x=0, u=1$
$\bg_white \int_{0}^{\frac{\pi}{4}}\frac{\sin{2x}}{4+\cos^{2}{x}}dx=\int_{1}^{\frac{1}{2}}\frac{-du}{4+u}$
When you remove the negative sign you flip the boundaries.
$\bg_white \int_{1}^{\frac{1}{2}}\frac{-du}{4+u}=\int_{\frac{1}{2}}^{1}\frac{du}{4+u}$
$\bg_white \int_{\frac{1}{2}}^{1}\frac{du}{4+u}=\left[ln\left(4+u\right)\right]_{\frac{1}{2}}^{1}=ln\left(\5\right)-ln\left(4\frac{1}{2}\right)=ln\left(\frac{10}{9}\right)$

The next one requires you to master trigonometric identities
$\bg_white \int_{0}^{\frac{\pi}{2}}\cos{5x}\sin{3x}dx$
Have a look at this when you see opposites you already could tell it will be in the form of $\bg_white \frac{1}{2}\left(\sin\left(A+B\right)\pm{\sin\left(A-B\right)}\right)=\cos{A}\sin{B} \& \sin{A}\cos{B}$ depending on what is used.

Here, we are hunting for $\bg_white \cos{A}\sin{B}$ therefore, we will need something like this
$\bg_white \frac{1}{2}\left(\sin\left(A+B\right)-\sin\left(A-B\right)\right)$.
As a matter of fact $\bg_white A=5x \& B=3x$.
Thus, $\bg_white \int_{0}^{\frac{\pi}{2}}\cos{5x}\sin{3x}dx=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin{8x}-\sin{2x}dx$
$\bg_white \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin{8x}-\sin{2x}dx=\frac{1}{2}\left[\frac{-\cos{8x}}{8}+\frac{\cos{2x}}{2}\right]_{0}^{\frac{\pi}{2}}$
$\bg_white \frac{1}{2}\left[\frac{-1}{8}+\frac{-1}{2}\right]-\frac{1}{2}\left[\frac{-1}{8}+\frac{1}{2}\right]=\frac{-1}{2}$

i) $\bg_white \frac{d}{d\theta}\left(\sin^{3}{\theta}\right)=3\cos{\theta}\sin^{2}{\theta}$

ii)Use $\bg_white x=\tan{\theta}$ to evaluate $\bg_white \int_{0}^{1}\frac{x^{2}}{\left(1+x^{2}\right)^{\frac{5}{2}}}dx$
To start $\bg_white x=\tan{\theta} \& dx=\sec^{2}{\theta}d\theta$
When $\bg_white x=1\leftrightarrow{\theta=\frac{\pi}{4}}$ and when $\bg_white x=0\leftrightarrow{\theta=0}$
$\bg_white \int_{0}^{1}\frac{x^{2}}{\left(1+x^{2}\right)^{\frac{5}{2}}}dx=\int_{0}^{\frac{\pi}{4}}\frac{\tan^{2}{\theta}}{\sec^{3}{\theta}}d\theta$
$\bg_white \int_{0}^{\frac{\pi}{4}}\frac{\tan^{2}{\theta}}{\sec^{3}{\theta}}d\theta=\int_{0}^{\frac{\pi}{4}}\tan^{2}{\theta}\cos^{3}{\theta}d\theta=\int_{0}^{\frac{\pi}{4}}\sin^{2}{\theta}\cos{\theta}d\theta$.

Using part i it becomes $\bg_white \left[\frac{\sin^{3}{\theta}}{3}\right]_{0}^{\frac{\pi}{4}}=\frac{\frac{1}{2\sqrt{2}}}{3}-0=\frac{1}{6\sqrt{3}}$.

The question you saw here was the 3U version. You can turn this Q into a 4U question by removing the first part of the question and not giving you the substitution for the second part.

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