Integrate the following functions w.r.t. $x$: (i) $\sin mx$ (ii) $2x \sin(x^2 + 1)$ (iii) $\dfrac{ an^4 \sqrt{x}\,\sec^2 \sqrt{x}}{\sqrt{x}}$ (iv) $\dfrac{\sin( an^{-1} x)}{1 + x^2}$

Question

Integrate the following functions w.r.t. $x$: (i) $\sin mx$ (ii) $2x \sin(x^2 + 1)$ (iii) $\dfrac{	an^4 \sqrt{x}\,\sec^2 \sqrt{x}}{\sqrt{x}}$ (iv) $\dfrac{\sin(	an^{-1} x)}{1 + x^2}$

Accepted Answer

> [!TLDR]
> All four are u-substitutions: (i) $-\dfrac{\cos mx}{m}+C$;
> (ii) $-\cos(x^2+1)+C$;
> (iii) $\dfrac{2}{5}	an^5\sqrt{x}+C$;
> (iv) $-\cos(	an^{-1}x)+C=-\dfrac{1}{\sqrt{1+x^2}}+C$.

## 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. $dx=\dfrac{du}{m}$:
$$\int\sin mx\,dx=\frac{1}{m}\int\sin u\,du=-\frac{1}{m}\cos u+C=-\frac{\cos mx}{m}+C.$$
Check: $\dfrac{d}{dx}\left(-\dfrac{\cos mx}{m}ight)=\sin mx.$

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

Here $u=x^2+1$ has $du=2x\,dx$ — exactly the factor present:
$$\int\sin u\,du=-\cos u+C=-\cos(x^2+1)+C.$$

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

Take $u=	an\sqrt{x}$. Then
$$du=\sec^2\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,dx\quad\Rightarrow\quad \frac{\sec^2\sqrt{x}}{\sqrt{x}}\,dx=2\,du.$$
The integrand is $	an^4\sqrt{x}\cdot\dfrac{\sec^2\sqrt{x}}{\sqrt{x}}\,dx=u^4\cdot 2\,du$, so
$$\int 2u^4\,du=\frac{2u^5}{5}+C=\frac{2}{5}	an^5\sqrt{x}+C.$$

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

Let $u=	an^{-1}x$, so $du=\dfrac{dx}{1+x^2}$:
$$\int\sin u\,du=-\cos u+C=-\cos(	an^{-1}x)+C.$$
A right triangle with opposite $x$, adjacent $1$, hypotenuse $\sqrt{1+x^2}$ gives $\cos(	an^{-1}x)=\dfrac{1}{\sqrt{1+x^2}}$, so this is also $-\dfrac{1}{\sqrt{1+x^2}}+C$.

> [!ANSWER]
> (i) $-\dfrac{\cos mx}{m}+C$;
> (ii) $-\cos(x^2+1)+C$;
> (iii) $\dfrac{2}{5}	an^5\sqrt{x}+C$;
> (iv) $-\cos(	an^{-1}x)+C=-\dfrac{1}{\sqrt{1+x^2}}+C$

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

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?

The Definite Integral Case

Common Mistake to Avoid

The common idea

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

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

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

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

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

Steps

Common Mistakes

Q.Integrate the following functions w.r.t. xxx:

The common idea

(i) ∫sin⁡mx dx\int\sin mx\,dx∫sinmxdx

(ii) ∫2xsin⁡(x2+1) dx\int 2x\sin(x^2+1)\,dx∫2xsin(x2+1)dx

(iii) ∫tan⁡4x sec⁡2xx dx\int\dfrac{\tan^4\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\,dx∫x​tan4x​sec2x​​dx

(iv) ∫sin⁡(tan⁡−1x)1+x2 dx\int\dfrac{\sin(\tan^{-1}x)}{1+x^2}\,dx∫1+x2sin(tan−1x)​dx

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

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

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

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

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

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