Prove that $\int_{0}^{\pi/2}\sin^3 x\,dx=\frac{2}{3}$

Prove that ∫₀π/2sin³ xdx=(2)/(3)

Miscellaneous Exercise · Q35

\int_{0}^{\pi/2}\sin^3 x\,dx=\frac{2}{3}

Puducherry CbseNCERTSubjective· 3mImportance★★★★★

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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The integral $\int_{0}^{\pi/2}\sin^3 x\,dx$ is evaluated by rewriting $\sin^3 x$ as $\sin x (1 - \cos^2 x)$ and using the substitution $u = \cos x$ , which transforms the integral into a simple polynomial form. The value is $\frac{2}{3}$ .

The key to evaluating powers of sine or cosine over a symmetric interval like $[0, \pi/2]$ is often to use a trigonometric identity to reduce the power, then substitute. For $\sin^3 x$ , the direct approach is to factor it as $\sin x \cdot \sin^2 x$ , then replace $\sin^2 x$ with $1 - \cos^2 x$ . This sets up a perfect substitution because the derivative of $\cos x$ is $-\sin x$ , which appears as a factor.

Let’s work through it step by step.

Rewrite the integrand We have $\sin^3 x = \sin x \cdot \sin^2 x = \sin x (1 - \cos^2 x)$ . So the integral becomes

I = \int_{0}^{\pi/2} \sin x (1 - \cos^2 x) \, dx.

Choose a substitution

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

When $x = 0$ , $u = \cos 0 = 1$ . When $x = \pi/2$ , $u = \cos(\pi/2) = 0$ .

The limits reverse: the lower limit becomes $u=1$ and the upper limit becomes $u=0$ .
Transform the integral

Substituting everything:

I = \int_{x=0}^{x=\pi/2} \underbrace{(1 - \cos^2 x)}_{1 - u^2} \cdot \underbrace{\sin x \, dx}_{-du} = \int_{u=1}^{u=0} (1 - u^2)(-du).

The minus sign flips the limits:

$I = \int_{0}^{1} (1 - u^2) \, du.$ …

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Q.Prove that $\int_{0}^{\pi/2}\sin^3 x\,dx=\frac{2}{3}$

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

Method: Odd power of sine — peel one factor and substitute $u=\cos x$

Steps

Common Mistakes

Q.Prove that ∫0π/2sin⁡3x dx=23\int_{0}^{\pi/2}\sin^3 x\,dx=\frac{2}{3}∫0π/2​sin3xdx=32​

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Q.Prove that $\int_{0}^{\pi/2}\sin^3 x\,dx=\frac{2}{3}$

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