If _0^(3x)e^-^2x,dx=-e_0^1t,e^t,dt, then + is equal to _____ .

Question

If $\displaystyle\int_0^\pi(\sin3x)e^{-\sin^2x}\,dx=\alpha-\dfrac{\beta}{e}\displaystyle\int_0^1\sqrt{t}\,e^t\,dt$, then $\alpha+\beta$ is equal to \_\_\_\_\_ .

Answer: 5

Accepted Answer

$$I=\int_0^\pi(\sin3x)e^{-\sin^2x}\,dx=2\int_0^{\pi/2}\sin x\,e^{-\sin^2x}dx+\int_0^{\pi/2}\cos x\,e^{-\sin^2x}(-\sin2x)\,dx$$

Integrating the second term by parts and simplifying:
$$I=3\int_0^{\pi/2}\sin x\,e^{-\sin^2x}dx-1$$

Put $-\sin^2x=\alpha'$ (auxiliary variable), then $1+\alpha'=x'$ (substitution chain) leading to
$$I=\frac{3}{2e}\int_0^1e^x\frac{1}{\sqrt x}dx-1=2-\frac{3}{e}\int_0^1e^x\sqrt x\,dx$$

Comparing with $\alpha-\dfrac\beta e\displaystyle\int_0^1\sqrt te^tdt$: $\alpha=2,\ \beta=3$, hence $\alpha+\beta=5$.

Solutions · 1