Q.Evaluate the integral using substitution $\int_{0}^{\pi/2}\frac{\sin x}{1+\cos^2 x}\,dx$

The key idea is U Substitution: choose $u = \cos x$, which simplifies the denominator and the differential.

Step 1: Let $u = \cos x$. Then $du = -\sin x \, dx$, so $\sin x \, dx = -du$.

Step 2: Change the limits: when $x = 0$, $u = \cos 0 = 1$; when $x = \pi/2$, $u = \cos(\pi/2) = 0$.

Step 3: Substitute into the integral:

$\int_{0}^{\pi/2}\frac{\sin x}{1+\cos^2 x}\,dx = \int_{1}^{0} \frac{-du}{1+u^2} = \int_{0}^{1} \frac{du}{1+u^2}.$ …

Why substitution works here

When you see $\sin x$ in the numerator and $\cos^2 x$ in the denominator, a natural instinct is to try $u = \cos x$. Why? Because the derivative of $\cos x$ is $-\sin x$, which almost exactly matches the $\sin x$ in the numerator — we just need a minus sign. This turns a trigonometric integral into a rational one, which is far easier to handle.

The limits also cooperate: when $x = 0$, $\cos 0 = 1$; when $x = \pi/2$, $\cos(\pi/2) = 0$. So the new limits become $u = 1$ to $u = 0$, which we can flip.

Step-by-step

1. Set up the substitution.

   Let $u = \cos x$. Then $du = -\sin x \, dx$, so $\sin x \, dx = -du$.

2. Rewrite the integral.

   The original integral is

Substituting, we get

3. Change the limits.
   • When $x = 0$, $u = \cos 0 = 1$.
   • When $x = \pi/2$, $u = \cos(\pi/2) = 0$. So the integral becomes

4. Flip the limits to remove the minus sign. Reversing the limits changes the sign:

(Alternatively, you can bring the minus inside: $-\int_{1}^{0} = \int_{0}^{1}$.)

5. Evaluate the standard integral. We know that

So …

Method: Substituting the trig function whose derivative sits in the numerator

For $\displaystyle\int \frac{\sin x}{1+\cos^2x}\,dx$-type integrals, substitute the function whose derivative is the numerator, turning a trig integral into a rational (often arctan) one.

Steps

Step 1: Choose $u$ so $du$ matches the numerator.

Here $u=\cos x$ gives $du=-\sin x\,dx$, so $\sin x\,dx=-du$ — the numerator is absorbed.

Step 2: Rewrite the denominator in $u$ and change the limits.

$1+\cos^2x$ becomes $1+u^2$; convert the $x$-limits to $u$-limits. …

Common Mistakes

Mistake 1: Dropping the minus sign from $du=-\sin x\,dx$.

Why it's wrong: $u=\cos x$ gives $\sin x\,dx=-du$; omitting the sign yields $-\tfrac{\pi}{4}$ instead of $+\tfrac{\pi}{4}$. Correct approach: carry the $-du$ and absorb it by reversing the limits.

Mistake 2: Choosing $u=\sin x$ instead of $u=\cos x$. …