-d x-sin 4 x-cos 4x is equal toA- 1-2tan -11-2tan 2 x - CB- None of these C- -2tan -11-2tan 2 x - CD- 1-2tan -11-2 2 x - C

The correct option is A 1/√(2) tan^ 1 ( 1/√(2)tan 2x )+C∫dx/sin^4 x +cos^4 x =∫dx/(sin^2 x+cos^2 x)^2 2 sin^2 x cos^2 x =∫2 dx/2 sin^2 2x=∫2 sec^2 2x/2 sec^ ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals -4

∫d x/sin 4 x+...

Question

$\int \frac{d x}{s i n^{4} x + c o s^{4} x}$ is equal to

\frac{1}{\sqrt{2}} t a n^{- 1} (\frac{1}{\sqrt{2}} t a n 2 x) + C

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\sqrt{2} t a n^{- 1} (\frac{1}{\sqrt{2}} t a n 2 x) + C

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\frac{1}{\sqrt{2}} t a n^{- 1} (\frac{1}{\sqrt{2}} c o t 2 x) + C

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None of these

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Solution

The correct option is A $\frac{1}{\sqrt{2}} t a n^{- 1} (\frac{1}{\sqrt{2}} t a n 2 x) + C$
$\int \frac{d x}{s i n^{4} x + c o s^{4} x} = \int \frac{d x}{(s i n^{2} x + c o s^{2} x)^{2} - 2 s i n^{2} x c o s^{2} x} = \int \frac{2 d x}{2 - s i n^{2} 2 x} = \int \frac{2 s e c^{2} 2 x}{2 s e c^{2} 2 x - t a n^{2} 2 x} d x = 2 \int \frac{s e c^{2} 2 x}{2 + t a n^{2} 2 x} d x = \int \frac{d t}{(\sqrt{2})^{2} + t^{2}} [p u t t i n g t a n 2 x = t \Rightarrow 2 s e c^{2} 2 x d x = d t] = \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{t}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{1}{\sqrt{2}} t a n 2 x) + C$

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∫dxsin4x+cos4x is equal to

The correct option is A 1√2tan−1(1√2tan 2x)+C∫dxsin4 x +cos4 x=∫dx(sin2 x+cos2 x)2−2 sin2 x cos2 x=∫2 dx2−sin2 2x=∫2 sec2 2x2 sec2 2x−tan2 2xdx=2∫sec2 2x2+tan2 2xdx=∫dt(√2)2+t2[putting tan 2x=t⇒2 sec2 2x dx=dt]=1√2tan−1(t√2)+C=1√2tan−1(1√2tan 2x)+C

$\int \frac{d x}{s i n^{4} x + c o s^{4} x}$ is equal to