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Calculation of $\displaystyle \int\frac{x^2 \left(\sin 2x - \cos 2x\right)}{\left(1+\sin 2x\right)\cos^2 x}dx$

$\bf{My\; Try}::$ We can write the given Integral as $\displaystyle \int\frac{x^2 \left\{(1+\sin 2x)-(1+\cos 2x)\right\}}{(1+\sin 2x)\cos^2 x}dx$

$\displaystyle = \int\frac{x^2}{\cos^2 x}dx - \int \frac{2x^2}{1+\sin 2x}dx$

$\displaystyle = \int x^2 \sec^2 xdx - \int\frac{x^2}{\cos^2 \left(\frac{\pi}{4}-x\right)}dx$

$\displaystyle = \int x^2 \sec^2 xdx - \int x^2 \sec^2 \left(\frac{\pi}{4}-x\right)dx$

Now I did not understand how can i solve that

Help Required


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Use with $x^2$ as the first term –  lab bhattacharjee Dec 12 '13 at 17:00
You may use Integration by Parts. –  Ahaan S. Rungta Dec 12 '13 at 17:00

2 Answers 2

You will need to use integration by parts. This is:

$\displaystyle \int u dv = uv - \int vdu$

For your problem it's a bit tedious, but doable... Make things like x^2 u because it will become a constant eventually.

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My hint

$$\int x^2d(\tan x)+\int x^2d(\tan(\frac{\pi}{4}-x))=x^2(\tan x+\tan(\frac{\pi}{4}-x))-2\int x(\tan x+\tan(\frac{\pi}{4}-x))dx$$

$$\int x(\tan x+\tan(\frac{\pi}{4}-x))dx=\int \frac{x}{\cos x(\sin x+\cos x)}dx$$

This is the result.

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