Calculate the integral $\int_{0}^{\frac{\pi}{2}} \cos(x) \left( \int_{1}^{\sin(x)} e^{t^2} dt \right) dx$

Question

Calculate the following integral:$\int_{0}^{\frac{\pi}{2}} \cos(x) \left( \int_{1}^{\sin(x)} e^{t^2} dt \right) dx$. I tried integrating by parts but ended up getting Gauss error function (which hasn't been contemplated by my course yet).Is there a way to calculate the integral without going through the gauss function?Any help would be kindly appreciated.

Answer 1

$$I=-\int_{0}^{\frac{\pi}{2}} \cos(x) \left( \int^{1}_{\sin(x)} e^{t^2} dt \right) dx$$

Switch the order of integration

$$I=-\int_{0}^{1} e^{t^2} \left(\int_0^{\arcsin(t)} \cos(x) dx\right)dt=-\int_{0}^{1} e^{t^2}\cdot t~dt=\frac{1-e}2$$

Answer 2

I can't comment yet, so a quick bug-fix to MathFail's answer. When you switch the order of integration, the variables lie on an $x$-$t$ plane with $0 \leq x \leq \frac{\pi}{2}$ above the sine function, so after switching the order, $t$ should span between $0$ and $1$, and $x$ between $0$ and $\arcsin(t)$.

As a sanity check, $t$ can never attain the value of $\frac{\pi}{2}$ in the original equation, it is bounded by $1$.

Answer 3

Integrate by parts

\begin{align} &\int_{0}^{\frac{\pi}{2}} \cos x \left( \int_{1}^{\sin2x} e^{t^2} dt \right) dx\\ =& \int_{0}^{\frac{\pi}{2}} \left( \int_{1}^{\sin x } e^{t^2} dt \right) d(\sin x) =-\int_0^{\frac\pi2}e^{\sin^2 x}\sin x\cos x\ \overset{y=2x}{dx}\\ =&\ \frac14 e^{\frac12}\int_0^\pi e^{-\frac12\cos y}\ d(\cos y) =\frac12(1-e) \end{align}