∫ 2x 4x/^4 x (1+² 2x) dx =

Correct Answer is : log ((1+cos⁡2x)21+cos⁡22x)+sec⁡2≤ft( (1+ 2x)²/1+² 2x ) + ²(1+cos22x(1+cos2x)2)+sec2 x + c

log ≤ft[ 1+² 2x/1+ 2x ] - ² x + c

log ≤ft( 1+² 2x/1+ 2x ) + ² x + c

log ≤ft( 1+ 2x/1+² 2x ) + ² x + c

log ≤ft( (1+ 2x)²/1+² 2x ) + ² x + c

Correct Answer is : log ((1+cos⁡2x)21+cos⁡22x)+sec⁡2≤ft( (1+ 2x)²/1+² 2x ) + ²(1+cos22x(1+cos2x)2)+sec2 x + c

Test Index

TS EAMCET 7 May 2018 Shift 1 Solved Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

∫

\frac{\cos 2x \sin 4x}{\cos^4 x (1+\cos^2 2x)}

dx =

Solution:

I = \int \frac{\cos 2x \sin 4x}{\cos^4 x (1+\cos^2 2x)} dx

= 4 \int \frac{\cos 2x (2 \sin 2x \cos 2x)}{(1+\cos 2x)^2 (1+\cos^2 2x)} dx

Put

\cos 2x = t

-2 \sin 2x dx = dt

I = -4 \int \frac{t^2}{(1+t)^2 (1+t^2)} dt

= -4 \int \left[ \frac{1}{2(1+t)^2} - \frac{1}{2(1+t)} \right.

+ \frac{t}{2(1+t^2)} dt

= -2 \left[ -\frac{1}{1+t} - \log(1+t) \right.

\left[+ \frac{1}{2} \log(1+t^2) \right] + c

= \frac{2}{1+\cos 2x} + 2 \log(1+\cos 2x)

- \log(1+\cos^2 2x) ) + c

= \log \left( \frac{(1+\cos 2x)^2}{1+\cos^2 2x} \right)

+ \sec^2 x + c

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