- dx -sin 4 x - cos 4 y is equal toA- 1-2tan -11-2tan 2 x - CB- -2tan -11-2tan 2 x - CC- 1-2tan -11-2 2 x - CD- None of these

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

Integration of Trigonometric Functions

∫ dx /sin 4 x...

Question

$\int \frac{dx}{{sin}^{4} x + {cos}^{4} x}$ 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 $\frac{1}{\sqrt{2}} t a n^{- 1} (\frac{1}{\sqrt{2}} t a n 2 x) + C$
$\int \frac{dx}{{sin}^{4} x + {cos}^{4} x} = \int \frac{dx}{({sin}^{2} x + {cos}^{2} x)^{2} - 2 {sin}^{2} x {cos}^{2} x} = \int \frac{2 dx}{2 - {sin}^{2} 2 x} = \int \frac{2 {sec}^{2} 2 x}{2 {sec}^{2} 2 x - {tan}^{2} 2 x} dx = 2 \int \frac{{sec}^{2} 2 x}{2 + {tan}^{2} 2 x} dx = \int \frac{dt}{(\sqrt{2})^{2} + t^{2}} [p u t t i n g tan 2 x = t \Rightarrow 2 {sec}^{2} 2 x dx = dt] = \frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{t}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C$

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Similar questions

\int \frac{d x}{s i n^{4} x + c o s^{4} x}

is equal to

\int \frac{d x}{2 + s i n 2 x + c o s 2 x}

equals to

Q. Let

I_{1} = \int \frac{d x}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}

and

I_{2} = \int \frac{d x}{e^{3 x} + 8 e^{x} + 4 e^{- x}}

, then value of

I = I_{1} - 2 I_{2}

is equal to

\int \frac{f (x) ϕ^{'} (x) + ϕ (x) f^{'} (x)}{(f (x) ϕ (x) + 1) \sqrt{f (x) ϕ (x) - 1}} d x =

y = f (x) = {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = {\begin{matrix} 2 {tan}^{- 1} x if x \geq 0 - 2 {tan}^{- 1} x if x < 0 \end{matrix}

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∫dxsin4x+cos4 x 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

$\int \frac{dx}{{sin}^{4} x + {cos}^{4} x}$ is equal to