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KCET · Maths · Indefinite Integration

The value of \(\int e^{\sin x} \sin 2 x d x\) is

  1. A \(2 e^{\sin x}(\sin x-1)+C\)
  2. B \(2 e^{\sin x}(\sin x+1)+C\)
  3. C \(2 e^{\sin x}(\cos x+1)+C\)
  4. D \(2 e^{\sin x}(\cos x-1)+C\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2 e^{\sin x}(\sin x-1)+C\)

Step-by-step Solution

Detailed explanation

We have, \(\int e^{\sin x} \sin 2 x d x\)
\(\begin{aligned}
&=\int e^{\sin x} \cdot(2 \sin x \cos x) d x \\
&=2 \int e^{\sin x} \sin x \cos x d x
\end{aligned}\)
let \(\sin x=t\)
\(\begin{aligned}
\Rightarrow \cos x d x &=d t=2 \int e^{t} \cdot t d t \\
&=2\left[t \int e^{t} d t-\int\left(\frac{d}{d t}(t) \int e^{t} d t\right) d t\right] \\
&=2\left[t \cdot e^{t}-\int e^{t} d t\right] \\
&=2\left[t \cdot e^{t}-e^{t}\right]+C \\
&=2\left[e^{t}(t-1)\right]+C \\
&=2\left[e^{\sin x}(\sin x-1)\right]+C \\
&=2 e^{\sin x}(\sin x-1)+C
\end{aligned}\)