∫ 2x · 4x/^4 x (1 + ² 2x) dx =

Correct Answer is : log⁡(1+cos⁡2x)2(1+cos⁡22x)+sec⁡2x+c (1+ 2x)²/(1+² 2x) + ² x + clog(1+cos22x)(1+cos2x)2+sec2x+c

log ≤ft( 1+ 2x/1+² 2x ) + ² x + c

log (1+ 2x)²/1+² x + sec x + c

log 1+² 2x/(1+² 2x)² + sec x + c

Correct Answer is : log⁡(1+cos⁡2x)2(1+cos⁡22x)+sec⁡2x+c (1+ 2x)²/(1+² 2x) + ² x + clog(1+cos22x)(1+cos2x)2+sec2x+c

Test Index

AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

∫

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

dx =

Solution:

Consider the expression,

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

Put

\cos 2x = t

then,

-2 \sin 2x \, dx = dt

Thus,

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

Split the above into partial fractions,

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

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

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

+ \frac{1}{2} \log(1+t^2) ]

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

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

= \sec^2 x + 2 \log(1+\cos 2x)

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

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

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