Integration (1 Viewer)

hmim · May 7, 2023

Can someone please show me how to do this.

unofficiallyred12 · May 7, 2023

change sin2x into (sinx+cosx)^2-1

Drongoski · May 7, 2023

Is this a possible answer??

$-ln[(sin x + cos x) + \sqrt{(sin x + cos x)^2 - 1}] + C$

unofficiallyred12 · May 7, 2023

yep looks legit

Drongoski · May 7, 2023

Assuming suggested answer is correct, here's my effort:

$\int \frac{1} {\sqrt{u^2 - a^2}}du = ln (u + \sqrt {u^2 - a^2}) + C \\ \\ sin 2x = 2sinxcosx = (sin x + cos x)^2 - 1 \\ \\ -d(sin x + cos x) = (cos x - sin x)dx \\ \\ \therefore I = \int \frac {sin x - cos x}{\sqrt {sin 2x}} dx = -\int \frac {d(sin x + cos x)}{\sqrt {(sin x + cos x)^2 - 1}}\\ \\ = -ln [(sin x + cos x) + \sqrt {(sinx + cos x)^2 - 1}] + C$

carrotsss · May 7, 2023

If you expand sin(2x) to 2cosxsinx and then simplify you can get sqrt(tanx) + 1/sqrt(tanx) but idk where to go from there

hmim · May 8, 2023

Drongoski said:
Assuming suggested answer is correct, here's my effort:

$\int \frac{1} {\sqrt{u^2 - a^2}}du = ln (u + \sqrt {u^2 - a^2}) + C \\ \\ sin 2x = 2sinxcosx = (sin x + cos x)^2 - 1 \\ \\ -d(sin x + cos x) = (cos x - sin x)dx \\ \\ \therefore I = \int \frac {sin x - cos x}{\sqrt {sin 2x}} dx = -\int \frac {d(sin x + cos x)}{\sqrt {(sin x + cos x)^2 - 1}}\\ \\ = -ln [(sin x + cos x) + \sqrt {(sinx + cos x)^2 - 1}] + C$

Yep, this is right. Thank you so much for your help!. The final answer just left the term inside the root as sin(2x) by expanding and simplifying

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Integration (1 Viewer)

hmim

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