Evaluate the definite integral $\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4 x+\sin^4 x}\,dx$

Evaluate the definite integral ∫₀π/4(sin xcos x)/(cos⁴ x+sin⁴ x)dx

Miscellaneous Exercise · Q25

\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4 x+\sin^4 x}\,dx

Punjab PsebTextbookSubjective· 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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Dividing by $\cos^4 x$ turns the integrand into $\dfrac{\tan x\sec^2 x}{1+\tan^4 x}$ ; with $t=\tan^2 x$ it becomes $\tfrac12\int_0^1\frac{dt}{1+t^2}=\dfrac{\pi}{8}$ .

The idea

The numerator and denominator are both built from powers of $\sin x$ and $\cos x$ . Dividing through by the highest power of $\cos x$ rewrites everything in terms of $\tan x$ and $\sec^2 x$ — and $\sec^2 x$ is exactly the derivative a $\tan$ -substitution needs.

Set up

Divide numerator and denominator by $\cos^4 x$ :

\frac{\sin x\cos x}{\cos^4 x+\sin^4 x}=\frac{\dfrac{\sin x\cos x}{\cos^4 x}}{1+\dfrac{\sin^4 x}{\cos^4 x}}=\frac{\tan x\sec^2 x}{1+\tan^4 x},

since $\dfrac{\sin x\cos x}{\cos^4 x}=\dfrac{\sin x}{\cos^3 x}=\tan x\sec^2 x$ .

Substitute

Let $t=\tan^2 x$ . Then

$\frac{dt}{dx}=2\tan x\sec^2 x\quad\Rightarrow\quad \tan x\sec^2 x\,dx=\tfrac12\,dt.$ …

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Q.Evaluate the definite integral $\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4 x+\sin^4 x}\,dx$

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?

The idea

Set up

Substitute

Method: Divide by $\cos^{k}x$ to convert into a $\tan x$ substitution

Steps

Common Mistakes

Q.Evaluate the definite integral ∫0π/4sin⁡xcos⁡xcos⁡4x+sin⁡4x dx\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4 x+\sin^4 x}\,dx∫0π/4​cos4x+sin4xsinxcosx​dx

The idea

Set up

Substitute

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Q.Evaluate the definite integral $\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4 x+\sin^4 x}\,dx$

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