Q.Prove that $\int_{0}^{\pi/4}2\tan^3 x\,dx=1-\log 2$

The key idea is to rewrite $\tan^3 x$ as $\tan x \cdot \tan^2 x$, then use the identity $\tan^2 x = \sec^2 x - 1$ and a U Substitution.

Step 1: Write the integral and split the integrand.

Step 2: Separate into two integrals.

Step 3: For the first integral, let $u = \tan x$, so $du = \sec^2 x \, dx$. When $x=0$, $u=0$; when $x=\pi/4$, $u=1$. The second integral is standard: $\int \tan x \, dx = -\log|\cos x|$.

Step 4: Evaluate. …

Why This Approach Works

The integrand $2\tan^3 x$ looks tricky because $\tan^3 x$ doesn't have a simple antiderivative directly. But here's the key insight: $\tan^3 x = \tan x \cdot \tan^2 x$, and $\tan^2 x = \sec^2 x - 1$ (from the Pythagorean identity). This rewrites the integral into two pieces:

• One part involves $\tan x \sec^2 x$, which is a perfect candidate for $u$-substitution with $u = \tan x$ (since $du = \sec^2 x\,dx$).
• The other part is just $\tan x$, whose antiderivative is $-\log|\cos x|$.

This decomposition turns a messy trigonometric integral into clean, elementary forms.

Step-by-Step Solution

1. Rewrite the integrand using the identity.

Recall: $\tan^2 x = \sec^2 x - 1$. So:

Thus the integral becomes:

2. Handle the first integral: $I_1 = \int 2\tan x \sec^2 x\,dx$.

Let $u = \tan x$. Then $du = \sec^2 x\,dx$. The factor $2$ is constant, so:

No constant needed since we'll evaluate definite limits. So:

3. Handle the second integral: $I_2 = \int 2\tan x\,dx$.

We know $\int \tan x\,dx = -\log|\cos x| + C$. So:

$\int 2\tan x\,dx = -2\log|\cos x| + C.$ …

Method: Reduce $\tan^{3}x$ with $\tan^{2}x=\sec^{2}x-1$

Use this for odd/high powers of tangent: peel a $\tan^2x$ and replace it by $\sec^2x-1$ to split into a substitution piece and a plain $\tan x$ piece.

Steps

Step 1: Rewrite using the Pythagorean identity.

$\tan^3x=\tan x\,(\sec^2x-1)=\tan x\sec^2x-\tan x$.

Step 2: Integrate the $\tan x\sec^2x$ piece. …

Common Mistakes

Mistake 1: Trying to integrate $\tan^3x$ directly.

Why it's wrong: there is no simple direct antiderivative; the identity $\tan^2x=\sec^2x-1$ is needed to split it. Correct approach: write $\tan^3x=\tan x\sec^2x-\tan x$.

Mistake 2: Sign error in $\int\tan x\,dx$.

Why it's wrong: $\int\tan x\,dx=-\log|\cos x|$; dropping the minus flips the $\log 2$ term. Correct approach: keep the negative log-cosine. …