Q.Evaluate: $\int_{0}^{\pi/2} \dfrac{\tan x}{1+m^2\tan^2 x}\,dx$

The key idea is to rewrite the integrand in terms of $\sin x$ and $\cos x$, then use a substitution that converts the integral into a standard arctangent form.

First, write $\tan x = \frac{\sin x}{\cos x}$:

Multiply numerator and denominator by 2, and use $\sin 2x = 2\sin x \cos x$:

Now substitute $t = \cos^2 x + m^2 \sin^2 x$. Then $dt = (-2\cos x \sin x + 2m^2 \sin x \cos x)\,dx = (m^2 - 1) \sin 2x \,dx$. When $x=0$, $t=1$; when $x=\pi/2$, $t=m^2$. Hence: …

1. Rewrite the integrand. With $\tan x = \dfrac{\sin x}{\cos x}$ and multiplying top and bottom by $\cos^2 x$,

Since $\cos^2 x = 1-\sin^2 x$, the denominator becomes $1+(m^2-1)\sin^2 x$:

2. Substitute $t=\sin^2 x$. Then $dt = 2\sin x\cos x\,dx$, and the limits run $t:0\to 1$:

3. Evaluate for $m^2\neq 1$.

$I = \frac{1}{2(m^2-1)}\Big[\log\big|1+(m^2-1)t\big|\Big]_{0}^{1} = \frac{\log(m^2)}{2(m^2-1)}.$ …

Method: Reducing a trig-rational definite integral by substitution

Use this for $\int_0^{\pi/2}\dfrac{\tan x}{1+m^2\tan^2 x}\,dx$-type integrals. Rewriting in $\sin,\cos$ and substituting $u=\sin^2 x$ (or $\cos 2x$) turns it into a simple rational integral.

Steps

Step 1: Rewrite $\tan x$ in terms of $\sin x,\cos x$.

Step 2: Substitute $u=\sin^2 x$.

The integral becomes $\tfrac12\displaystyle\int_0^1\frac{du}{(1-u)+m^2u}$.

Step 3: Integrate the linear-denominator form. …

Common Mistakes

Mistake 1: Ignoring the $m^2=1$ special case.

Why it's wrong: the general formula $\tfrac{\log m^2}{2(m^2-1)}$ divides by $m^2-1$, undefined at $m^2=1$; there the integrand is $\tfrac12\sin 2x/(\ldots)$ giving $\tfrac12$. Correct approach: state the value at $m^2=1$ separately.

Mistake 2: Not substituting $u=\sin^2 x$. …