MHT CET · Maths · Indefinite Integration
\(\int \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x=\)
- A \(\tan ^{-1}\left(\sin ^{2} x\right)+c\)
- B \(2 \tan ^{-1}\left(\tan ^{2} x\right)+c\)
- C \(\frac{1}{2} \tan ^{-1}\left(\tan ^{2} x\right)+c\)
- D \(\tan ^{-1}\left(\cos ^{2} x\right)+c\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2} \tan ^{-1}\left(\tan ^{2} x\right)+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x\)
Dividing numerator and denominator by \(\cos ^{4} x\), we get
\(
I=\int \frac{\tan x \sec ^{2} x}{\tan ^{4} x+1} d x
\)
Put \(\tan ^{2} x=t \Rightarrow 2 \tan x \sec ^{2} x d x=d t\)
\(
I=\frac{1}{2} \int \frac{d t}{1+t^{2}}=\frac{1}{2} \tan ^{-1} t=\frac{1}{2} \tan ^{-1}\left(\tan ^{2} x\right)+c
\)
Dividing numerator and denominator by \(\cos ^{4} x\), we get
\(
I=\int \frac{\tan x \sec ^{2} x}{\tan ^{4} x+1} d x
\)
Put \(\tan ^{2} x=t \Rightarrow 2 \tan x \sec ^{2} x d x=d t\)
\(
I=\frac{1}{2} \int \frac{d t}{1+t^{2}}=\frac{1}{2} \tan ^{-1} t=\frac{1}{2} \tan ^{-1}\left(\tan ^{2} x\right)+c
\)
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