Integrate the following function: sin x sin (cos x)

Exercise 7.2 · Q4

\sin x \sin (\cos x)

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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 key idea is to recognise that $\cos x$ sits inside the outer sine, and its derivative $-\sin x$ is right there as a factor — this is a textbook case for u-substitution. Letting $u = \cos x$ transforms the integral into $\int -\sin u \, du$ , which integrates directly to $\cos u + C$ . Substituting back gives the final answer: $\boxed{\cos(\cos x) + C}$ .

Why substitution works here

When you see a function of the form $f(g(x)) \cdot g'(x)$ , substitution is almost always the cleanest path. Here, the outer function is $\sin(\text{something})$ , the inner function is $\cos x$ , and the derivative of the inner function is $-\sin x$ — which is exactly the factor sitting next to the integral sign (up to a sign). That’s the signal: let $u = \cos x$ .

Step-by-step

Set up the substitution. Let $u = \cos x$ . Then differentiate:

\frac{du}{dx} = -\sin x \quad\Rightarrow\quad du = -\sin x \, dx.

Rewrite the integral in terms of $u$ . The original integral is $\int \sin x \, \sin(\cos x) \, dx$ . Notice that $\sin x \, dx$ appears — but we have $du = -\sin x \, dx$ , so $\sin x \, dx = -du$ . Substituting:

\int \sin x \, \sin(\cos x) \, dx = \int \sin(u) \, (-du) = -\int \sin u \, du.

Integrate with respect to $u$ . The integral of $\sin u$ is $-\cos u$ , so: …

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Q.Integrate the following function: $\sin x \sin (\cos x)$

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?

Why substitution works here

Step-by-step

Method: Substitute the argument of a composite trig function

Steps

Common Mistakes

Q.Integrate the following function: sin⁡xsin⁡(cos⁡x)\sin x \sin (\cos x)sinxsin(cosx)

Why substitution works here

Step-by-step

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Q.Integrate the following function: $\sin x \sin (\cos x)$

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