MHT CET · Maths · Indefinite Integration
The value of \(\int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \mathrm{~d} x\) is
- A \(\quad \log (\sin x)-\sin x+\mathrm{c}\), where c is a constant of integration.
- B \(\log (\sin x)-\cos x+\mathrm{c}\), where c is a constant of integration.
- C \(\log (\sin x)+\sin x+\mathrm{c}\), where c is a constant of integration.
- D \(\log (\cos x)-\cos x+\mathrm{c}\), where c is a constant of integration.
Answer & Solution
Correct Answer
(A) \(\quad \log (\sin x)-\sin x+\mathrm{c}\), where c is a constant of integration.
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& I=\int \frac{\left(1-\sin ^2 x\right) \cos x}{\sin x(1+\sin x)} d x \\
& =\int \frac{1-\sin x}{\sin x} \cos x d x
\end{aligned}\)
Put \(t=\sin x\)
\(\Rightarrow \cos x d x=d t\)
So, \(I=\int\left(\frac{1}{t}-1\right) d t\)
\(\begin{aligned}
& =\log t-t+C \\
& \Rightarrow I=\log \sin x-\sin x+C
\end{aligned}\)
& I=\int \frac{\left(1-\sin ^2 x\right) \cos x}{\sin x(1+\sin x)} d x \\
& =\int \frac{1-\sin x}{\sin x} \cos x d x
\end{aligned}\)
Put \(t=\sin x\)
\(\Rightarrow \cos x d x=d t\)
So, \(I=\int\left(\frac{1}{t}-1\right) d t\)
\(\begin{aligned}
& =\log t-t+C \\
& \Rightarrow I=\log \sin x-\sin x+C
\end{aligned}\)
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