Q.Evaluate : $\int_{0}^{\pi/4} \dfrac{\sin x + \cos x}{16 + 9\sin 2x}\, dx$ OR Evaluate : $\int_{1}^{3} (x^2 + 3x + e^x)\, dx$ as the limit of the sum.

With $f(x)=\dfrac{1}{20}(x-5)^2$:

$\int_{-5}^{5}f(x)\,dx=\frac{1}{20}\cdot\frac{(x-5)^3}{3}\Big|_{-5}^{5}=\frac{1}{60}\big[0-(-10)^3\big]=\frac{1000}{60}=\frac{50}{3}.$ …

Let $u=t+1$, so $t=u-1$, $dt=du$, and limits $t:0\to1$ become $u:1\to2$:

$\int_1^2\dfrac{u-1}{u^3}du=\int_1^2\bigl(u^{-2}-u^{-3}\bigr)du=\left[-\dfrac1u+\dfrac{1}{2u^2}\right]_1^2.$ …

Note that

so the integrand equals $(t+3)^3+2+(t+3)\cos(t+3)$.

Substitute $u=t+3$ (so $t=-6\Rightarrow u=-3$ and $t=0\Rightarrow u=3$):

$\int_{-3}^{3}\left[u^3+2+u\cos u\right]du.$ …

Let $u=x^{16}$, $du=16x^{15}\,dx$. Then $1+x^{32}=1+u^2$ and $\cos(\tan^{-1}x^{16})=\dfrac{1}{\sqrt{1+u^2}}$.

$I=\displaystyle\int_0^1\dfrac{x^{15}}{1+x^{32}}\cos(\tan^{-1}x^{16})\,dx=\dfrac{1}{16}\int_0^1\dfrac{du}{(1+u^2)^{3/2}}$. …

Let $I=\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}$. Replace $x\to \tfrac{\pi}{6}+\tfrac{\pi}{3}-x=\tfrac{\pi}{2}-x$; then $\tan(\tfrac{\pi}{2}-x)=\cot x$, so

$I=\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot x}}=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt{\tan x}\,dx}{\sqrt{\tan x}+1}$. …

$\int_0^\pi\frac{\sin x}{1+\sin x}dx=\int_0^\pi\Big(1-\frac{1}{1+\sin x}\Big)dx=\pi-\int_0^\pi\frac{dx}{1+\sin x}$.

With $t=\tan(x/2)$, $1+\sin x=\frac{(1+t)^2}{1+t^2}$ and $dx=\frac{2\,dt}{1+t^2}$, so $\frac{dx}{1+\sin x}=\frac{2\,dt}{(1+t)^2}$. …

Between 3pm ($t=0$) and 5pm ($t=2$), the distance covered equals

Integrate term by term: …

Concept and Intuition

Write $\cot x/\sin x$ as $\cos x/\sin^2x$ so the substitution $u=\sin x$ applies directly.

Step-by-Step Solution

1. $\frac{\cot x}{\sin x}=\frac{\cos x}{\sin x}\cdot\frac1{\sin x}=\frac{\cos x}{\sin^2x}$.
2. Let $u=\sin x$, $du=\cos x\,dx$: antiderivative $=\int u^{-2}\,du=-\frac1u=-\frac1{\sin x}$.
3. Evaluate: $\left[-\frac1{\sin x}\right]_{\pi/6}^{\pi/2}=-\frac1{\sin(\pi/2)}+\frac1{\sin(\pi/6)}$. …

Concept and Intuition

The numerator $x^2$ is proportional to the derivative of $x^3+1$, so a substitution turns the integral into a simple power of $u$.

Step-by-Step Solution

1. Let $u=x^3+1$, $du=3x^2\,dx$; limits: $x=0\Rightarrow u=1$, $x=2\Rightarrow u=9$.
2. Integral $=\tfrac13\int_1^9 u^{-2}\,du=\tfrac13\big[-u^{-1}\big]_1^9$. …