Question

$\int e^{si{n}^{2}x}(cosx+co{s}^{3}x)sinxdx$ =

• A. $2\left[e^{si{n}^{2}x}\right(3-si{n}^{2}x\left)\right]+C$
• B. $\frac{1}{2}\left[e^{si{n}^{2}x}\right(3-si{n}^{2}x\left)\right]+C$
• C. $2\left[e^{si{n}^{2}x}\right(3+si{n}^{2}x\left)\right]+C$
• D. $\frac{1}{2}\left[e^{si{n}^{2}x}\right(3+si{n}^{2}x\left)\right]+C$

Answer 1

The correct option is A $2\left[e^{si{n}^{2}x}\right(3-si{n}^{2}x\left)\right]+C$

$e^{{\sin }^{2x}}(\cos x+{\cos }^{3}x)\sin xdx$ $\cos 2x=1-{\sin }^{2}x$

$\int e^{{\sin }^{2x}}(1+{\cos }^{2}x)\frac{2\sin x\cos x}{2}dx$ $2{\sin }^{2}x=1-\cos 2x$

$\int \frac{e\frac{(1-\cos 2x)}{2}}{2}(1+{\cos }^{2}x)\left(\sin 2xdx\right)\Rightarrow$ $1-\cos 2x=tt2\sin 2x=\frac{dt}{dx}$

$\frac{e^t}{2}\frac{(1+{\cos }^{2}x)}{2}dx=\int \frac{e^t}{2}\frac{\left(t\right)}{4}dt$ $1-2{\cos }^{2}x+1-t2(1-{\cos }^{2}x)=t$

$2\left[e^{{\sin }^{2}x}\right(3-{\sin }^{2}x\left)\right]+c$