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  • The value of the integral <img alt="\int e^{\sin ^2 x}\left(\cos x+\cos ^3 x\right) \sin x d x" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cint%20e%5E%7B%5Csin%20%5E2%20x%7D%5Cleft

The value of the integral \int e^{\sin ^2 x}\left(\cos x+\cos ^3 x\right) \sin x d x is equal to 

Option: 1

\frac{1}{2} e^{\sin ^2 x}\left(3-\sin ^2 x\right)+C


Option: 2

e^{\sin ^2 x}\left(1+\frac{1}{2} \cos ^2 x\right)+C


Option: 3

e^{\sin ^2 x}\left(3 \cos ^2 x+2 \sin ^2 x\right)+C


Option: 4

e^{\sin ^2 x}\left(2 \cos ^2 x+3 \sin ^2 x\right)+C


Answers (1)

best_answer

(a,b) Given integral is 

          \begin{aligned} & I=\int e^{\sin ^2 x}\left(\cos x+\cos ^3 x\right) \sin x d x \\ \\& \Rightarrow \quad I=\frac{1}{2} \int e^{\sin ^2 x}\left(1+\cos ^2 x\right) 2 \cos x \sin x d x \\ & \end{aligned}

On substituting t=\sin ^2 x \text { and } 2 \sin x \cos x d x=d t , we get

             \begin{aligned} I & =\frac{1}{2} \int e^{\sin ^2 x}\left(1+1-\sin ^2 x\right)(2 \cos x \sin x d x) \\ \\I & =\frac{1}{2} \int e^t(2-t) d t=\frac{3}{2} e^t-\frac{t e^t}{2}+C \\ \\& =\frac{1}{2} e^{\sin ^2 x}\left(3-\sin ^2 x\right)+C \\ \\& =e^{\sin ^2 x}\left(1+\frac{1}{2} \cos ^2 x\right)+C \end{aligned}

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jitender.kumar

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