∫ dx sin 4 x + cos 4 x is equal to Options: A . &nbsp 1 √ 2 tan − 1 ( 1 √ 2 tan 2 x ) + C B . &nbsp √ 2 tan − 1 ( 1 √ 2 tan 2 x ) + C C . &nbsp 1 √ 2 tan − 1 ( 1 √ 2 cot 2 x ) + C D . &nbsp None of these

Answer: Option A : A ∫ 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 2 x = ∫ 2 sec 2 2 x 2 sec 2 2 x − tan 2 2 x dx = 2 ∫ sec 2 2 x 2 + tan 2 2 x dx = ∫ dt ( √ 2 ) 2 + t 2 [ p u t t i n g tan 2 x = t ⇒ 2 sec 2 2 x dx = dt ] = 1 √ 2 tan − 1 ( t √ 2 ) + C = 1 √ 2 tan − 1 ( 1 √ 2 tan 2 x ) + C

∫dxsin4x+cos4 x is equal to

Discussion Forum : Indefinite Integration

Question -

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

Options:

A .

\frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C

B .

\sqrt{2} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C

C .

\frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} cot 2 x) + C

D . None of these

Answer: Option A
:
A

\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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Discussion Forum : Indefinite Integration

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