Question
Evaluate the following integrals:
$\int\cot^6\text{x}\text{ dx}$
$\int\cot^6\text{x}\text{ dx}$
✓
Answer
$\int\cot^6\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^4\text{x}\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}-\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int(\text{cosec}^2\text{x}-1)\text{ dx}$
Now, let $\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
And $\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
First we integrate I1
$\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
$\text{I}_1=-\int\text{t}^4\text{dt}+\int\text{t}^2\text{dt}$
$=\frac{-\text{t}^5}{5}+\frac{\text{t}^3}{3}+\text{C}_1$
$=-\frac{\cot^5\text{x}}{5}+\frac{\cot^3\text{x}}{3}+\text{C}_1$
Now we integrate I2
$\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
$=-\cot\text{x}-\text{x}+\text{C}_1$
Now, $\int\cot^6\text{x}\text{ dx}=\text{I}_1+\text{I}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{x}+\text{C}_1+\text{C}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{C}$ $\big[\therefore\text{ C}=\text{C}_1+\text{C}_2\big]$
$=\int\cot^4\text{x}\cdot\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^4\text{x}\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}-\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int(\text{cosec}^2\text{x}-1)\text{ dx}$
Now, let $\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
And $\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
First we integrate I1
$\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
$\text{I}_1=-\int\text{t}^4\text{dt}+\int\text{t}^2\text{dt}$
$=\frac{-\text{t}^5}{5}+\frac{\text{t}^3}{3}+\text{C}_1$
$=-\frac{\cot^5\text{x}}{5}+\frac{\cot^3\text{x}}{3}+\text{C}_1$
Now we integrate I2
$\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
$=-\cot\text{x}-\text{x}+\text{C}_1$
Now, $\int\cot^6\text{x}\text{ dx}=\text{I}_1+\text{I}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{x}+\text{C}_1+\text{C}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{C}$ $\big[\therefore\text{ C}=\text{C}_1+\text{C}_2\big]$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find $\frac{\text{dy}}{\text{dx}}$
y = ex + 10x + xx
→y = ex + 10x + xx
Differentiate $(\cos\text{x})^{\sin\text{x}}$ with respect to $(\sin\text{x})^{\cos\text{x}}$
→If four points A, B, C and D with position vectors $4\hat{\text{i}}+3\hat{\text{j}},5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}$ respectively are coplanar, then find the value of x.
→Show that $\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$ is discontinuous at x = 0.
→Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)
→$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=\log\text{x}$
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
→$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
If $\text{x}=\cos\text{t}+\log\tan\frac{\text{t}}{2},\text{y}=\sin\text{t},$ Then find the value of $\frac{\text{d}^2\text{y}}{\text{dt}^2}\ \text{and}\ \frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{4}.$
→In the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y.
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}-\text{x}\text{e}^\text{x}-\frac{5}{2}+\cos^2\text{x}$
→$\frac{\text{dy}}{\text{dx}}-\text{x}\text{e}^\text{x}-\frac{5}{2}+\cos^2\text{x}$
Find the value of k if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ where
$\text{f}\text{(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, &\text{ x}\neq\frac{\pi}{2}\\3, &\text{ x}=\frac{\pi}{2}\end{cases}$
→$\text{f}\text{(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, &\text{ x}\neq\frac{\pi}{2}\\3, &\text{ x}=\frac{\pi}{2}\end{cases}$