AP EAMCET · Maths · Indefinite Integration
\(k \in N, \int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x=\)
- A \(\frac{\tan x}{\sin ^{k+x}}+C\)
- B \(\frac{\tan x}{\sin ^k x}+C\)
- C \(\sin ^k x \sec ^2 x+C\)
- D \(k \sin ^{k-1} x \cos x+C\)
Answer & Solution
Correct Answer
(B) \(\frac{\tan x}{\sin ^k x}+C\)
Step-by-step Solution
Detailed explanation
\(I=\int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x\) \(\begin{aligned} & =\int \frac{\left(\sec ^2 x-k\right)}{\sin ^k x} d x \\ & =\int(\sin x)^{-k} \sec ^2 x d x-k \int \frac{d x}{\sin ^k x} \\ & =(\sin x)^{-k} \int \sec ^2 x d x-\end{aligned}\)…
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