Q.Integrate the following functions w.r.t. $x$:

Each integrand is an inner function times (a constant multiple of) its own derivative — a u-substitution.

(i) $u=mx,\ du=m\,dx$: $\displaystyle\int\sin mx\,dx=-\frac{1}{m}\cos mx+C.$

(ii) $u=x^2+1,\ du=2x\,dx$: $\displaystyle\int 2x\sin(x^2+1)\,dx=-\cos(x^2+1)+C.$

(iii) $u=\tan\sqrt{x}$, so $du=\dfrac{\sec^2\sqrt{x}}{2\sqrt{x}}\,dx$, giving $\dfrac{\sec^2\sqrt{x}}{\sqrt{x}}\,dx=2\,du$:

(iv) $u=\tan^{-1}x,\ du=\dfrac{dx}{1+x^2}$: $\displaystyle\int\frac{\sin(\tan^{-1}x)}{1+x^2}\,dx=-\cos(\tan^{-1}x)+C=-\frac{1}{\sqrt{1+x^2}}+C.$

The common idea

A u-substitution reverses the chain rule: if the integrand is $f(g(x))\,g'(x)$, set $u=g(x)$, $du=g'(x)\,dx$, and integrate $f(u)$. In each part, find the inner function whose derivative is present (perhaps up to a constant).

(i) $\int\sin mx\,dx$

Let $u=mx$, so $du=m\,dx$, i.e. $dx=\dfrac{du}{m}$:

Check: $\dfrac{d}{dx}\left(-\dfrac{\cos mx}{m}\right)=\sin mx.$

(ii) $\int 2x\sin(x^2+1)\,dx$

Here $u=x^2+1$ has $du=2x\,dx$ — exactly the factor present:

(iii) $\int\dfrac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\,dx$

Take $u=\tan\sqrt{x}$. Then

The integrand is $\tan^4\sqrt{x}\cdot\dfrac{\sec^2\sqrt{x}}{\sqrt{x}}\,dx=u^4\cdot 2\,du$, so

(iv) $\int\dfrac{\sin(\tan^{-1}x)}{1+x^2}\,dx$

Let $u=\tan^{-1}x$, so $du=\dfrac{dx}{1+x^2}$:

A right triangle with opposite $x$, adjacent $1$, hypotenuse $\sqrt{1+x^2}$ gives $\cos(\tan^{-1}x)=\dfrac{1}{\sqrt{1+x^2}}$, so this is also $-\dfrac{1}{\sqrt{1+x^2}}+C$.

Method: Reverse Chain Rule (Spotting $f(g(x))\,g'(x)$)

Use this for any integrand that is a composite function multiplied by (a constant times) the derivative of its inner part.

Steps

Step 1: Identify the inner function $g(x)$.

Look for a function whose derivative is present in the integrand. Candidates: the argument of a trig function ($mx$, $x^2+1$), or a nested expression such as $\tan\sqrt{x}$ or $\tan^{-1}x$.

Step 2: Set $u=g(x)$ and compute $du$.

Then $du=g'(x)\,dx$. Confirm the remaining factor in the integrand is $du$ up to a constant. For $u=mx$, $du=m\,dx$, so a $\frac1m$ is pulled out.

Step 3: Integrate in $u$ and restore $x$.

The integral reduces to a standard form in $u$ (e.g. $\int\sin u\,du=-\cos u$, $\int u^4\,du=\frac{u^5}{5}$). Finish by back-substituting $u=g(x)$ and adding $C$.

Common Mistakes

Mistake 1: Omitting the $\dfrac1m$ in $\int\sin mx\,dx$.

Why it's wrong: $du=m\,dx$ introduces a $\frac1m$; forgetting it gives $-\cos mx$ instead of $-\frac{\cos mx}{m}$. Correct approach: always divide by the constant from $du$.

Mistake 2: Missing the "hidden" $du$ in $\dfrac{\sec^2\sqrt{x}}{\sqrt{x}}$.

Why it's wrong: with $u=\tan\sqrt{x}$, $du=\frac{\sec^2\sqrt{x}}{2\sqrt{x}}\,dx$, so the whole factor is exactly $2\,du$. Correct approach: differentiate the composite inner function fully before deciding the substitution.

Mistake 3: Not simplifying $-\cos(\tan^{-1}x)$.

Why it's wrong: leaving it unsimplified hides the neat closed form $-\frac{1}{\sqrt{1+x^2}}$. Correct approach: use $\cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}}$.