TS EAMCET · Maths · Indefinite Integration
If \(I_n=\int \sin ^n x d x\), then \(n I_n-(n-1) I_{n-2}\) equals
- A \(\sin ^{n-1} x \cos x\)
- B \(\cos ^{n-1} x \sin x\)
- C \(-\sin ^{n-1} x \cos x\)
- D \(-\cos ^{n-1} x \sin x\)
Answer & Solution
Correct Answer
(C) \(-\sin ^{n-1} x \cos x\)
Step-by-step Solution
Detailed explanation
We know that, if \(\begin{aligned} & I_n=\int \sin ^n x d x \text {, then } \\ & I_n=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2} \end{aligned}\) where \(n\) is a positive integer. \(\Rightarrow \quad n I_n-(n-1) I_{n-2}=-\sin ^{n-1} x \cos x\)
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