JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int {\frac{{{{\sin }^2}\,x\,{{\cos }^2}\,x}}{{({{\sin }^3}\,x\, + {{\cos }^3}\,x)^2}}} dx\) is equal to
- A \(\frac{1}{{(1 + {{\cot }^3}\,x)}} + c\)
- B \( - \frac{1}{{3(1 + {{\tan }^3}\,x)}} + c\)
- C \(\frac{{{{\sin }^3}\,x}}{{(1 + {{\cos }^3}\,x)}} + c\)
- D \( - \frac{{{{\cos }^3}\,x}}{{3(1 + {{\sin }^3}\,x)}} + c\)
Answer & Solution
Correct Answer
(B) \( - \frac{1}{{3(1 + {{\tan }^3}\,x)}} + c\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{I}=\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x\) \(I=\int\left(\frac{\sin x \cdot \cos x}{\sin ^{3} x+\cos ^{3} x}\right)^{2} d x\)…
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