MHT CET · Maths · Indefinite Integration
The value of \(\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x\) is
- A \(\tan ^{-1}\left(\cot ^{2} x\right)+C\)
- B \(-\tan ^{-1}(\cos 2 x)+C\)
- C \(\tan ^{-1}(\sin 2 x)+C\)
- D \(\tan ^{-1}\left(\tan ^{2} x\right)+C\)
Answer & Solution
Correct Answer
(B) \(-\tan ^{-1}(\cos 2 x)+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x\)
\(I=\int \frac{\sin 2 x}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cdot \cos ^{2} x} d x\)
\(I=\int \frac{\sin 2 x}{1-\frac{1}{2}(\sin 2 x)^{2}} d x\)
\(\begin{aligned}=2 & \int \frac{\sin 2 x}{1+\left(1-\sin ^{2} 2 x\right)} d x \\ & \quad \text { (let } t=\cos 2 x \Rightarrow d t=-2 \sin 2 x d x \\=& \int \frac{-d t}{1+t^{2}} \\=&-\tan ^{-1} t+C \\=&-\tan ^{-1}(\cos 2 x)+C \end{aligned}\)
\(I=\int \frac{\sin 2 x}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cdot \cos ^{2} x} d x\)
\(I=\int \frac{\sin 2 x}{1-\frac{1}{2}(\sin 2 x)^{2}} d x\)
\(\begin{aligned}=2 & \int \frac{\sin 2 x}{1+\left(1-\sin ^{2} 2 x\right)} d x \\ & \quad \text { (let } t=\cos 2 x \Rightarrow d t=-2 \sin 2 x d x \\=& \int \frac{-d t}{1+t^{2}} \\=&-\tan ^{-1} t+C \\=&-\tan ^{-1}(\cos 2 x)+C \end{aligned}\)
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