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  • If     <img alt="I(x)=\int e^{\sin ^2 x}(\cos x \sin 2 x-\sin x)" src="https://entrancecorner.oncodecogs.com/gif.latex?I%28x%29%3D%5Cint%20e%5E%7B%5Csin%20%5E2%20x%7D%28%5Ccos%20x%20

If     I(x)=\int e^{\sin ^2 x}(\cos x \sin 2 x-\sin x)  dx and I (0) = 1, then   \mathrm{I}\left(\frac{\pi}{3}\right)     is equal to :

 

Option: 1

\mathrm{e}^{\frac{3}{4}}


Option: 2

\mathrm-{e}^{\frac{3}{4}}


Option: 3

\frac{1}{2}\mathrm{e}^{\frac{3}{4}}


Option: 4

-\frac{1}{2}\mathrm{e}^{\frac{3}{4}}


Answers (1)

best_answer

\begin{aligned} & I=\int \underbrace{e^{\sin ^2 x} \sin 2 x}_{11} \underbrace{\cos x d x}_1-\int e^{\sin ^2 x} \sin x d x \\ & =\cos x \int e^{\sin ^2 x} \sin 2 x d x-\int\left((-\sin x) \int e^{\sin ^2 x} \sin 2 x d x\right) d x-\int e^{\sin ^2 x} \sin x d x \\ & \sin ^2 \mathrm{x}=\mathrm{t} \\ & \sin 2 x d x=d t \\ \end{aligned}

\begin{aligned} & =\cos x \int e^t d t+\int\left(\sin x \int e^t d t\right) d x-\int e^{\sin ^2 x} \sin x d x \\ & =\mathrm{e}^{\sin ^2 x} \cos x+\int e^{\sin ^2 x} \sin x d x-\int e^{\sin ^2 x} \sin x d x \\ & \end{aligned}

\begin{aligned} & \mathrm{I}=\mathrm{e}^{\sin ^2 x} \cos \mathrm{x}+\mathrm{C} \\ & \mathrm{I}(0)=1 \\ & \Rightarrow 1=1+C \\ & \Rightarrow \mathrm{C}=0 \\ & \therefore \mathrm{I}=\mathrm{e}^{\sin ^2 x} \cos \mathrm{x} \\ & \mathrm{I}\left(\frac{\pi}{3}\right)=\mathrm{e}^{\sin ^2 \frac{\pi}{3}} \cos \frac{\pi}{3} \\ \end{aligned}

\begin{aligned} & =\frac{e^{\frac{3}{4}}}{2} \end{aligned}

 

Posted by

Shailly goel

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