Q31: $\int_0^{\pi/2}e^x\sin x\,dx$⁠, or $\int\frac{dx}{\cos(x - a)\cos(x - b)}$

The integrand is a product of a trig function and an exponential, so use integration by parts:

Take $u = \sin x$ and $v = e^x$⁠, so $\int v\,dx = e^x$ and $\frac{du}{dx} = \cos x$⁠.

Either choice of first function works, as long as the second step keeps the trig function first again.

The first step leaves $\int\cos x\,e^x\,dx$⁠. Integrate it by parts again with $\cos x$ first. Its derivative is $-\sin x$⁠.

Put this back into the first line:

That brings back $\int e^x\sin x\,dx$⁠, which is $I$ itself. Watch the signs: opening the bracket turns $-\int(-\sin x)e^x\,dx$ into $+\int(-\sin x)e^x\,dx$⁠, and that is $-I$⁠.

$I$ appears on both sides, so move it across: $2I = e^x\sin x - e^x\cos x$⁠. Then divide by 2.

No $+ C$ is written here, because this result is only used for the definite integral.

Put in the limits, upper minus lower, with

and $e^0 = 1$⁠. Watch the sign: $-\frac12(0 - 1) = +\frac12$⁠.

Take out $\frac12$ for the second form: $\frac12\left(e^{\pi/2} + 1\right)$⁠.

The trick is to multiply and divide by the constant $\sin(a - b)$⁠. Written as $\sin[(x - b) - (x - a)]$⁠, the top can be expanded into the same angles as the bottom.

This needs $\sin(a - b) \ne 0$⁠, which the question assumes.

Multiplying and dividing by the same number changes nothing: $\frac{\sin(a - b)}{\sin(a - b)} = 1$⁠.

Take $A = x - b$ and $B = x - a$⁠. Each part then cancels one cosine:

Split the fraction into two. In the first part $\cos(x - a)$ cancels; in the second part $\cos(x - b)$ cancels:

In the same way the second part gives $\tan(x - a)$⁠.

Standard integral:

With $u = x - b$ we have $du = dx$⁠, so

The same holds with $a$ in place of $b$⁠.