Question
Evaluate the following integrals:
$\int\cos^5\text{x}\text{ dx}$
$\int\cos^5\text{x}\text{ dx}$
✓
Answer
$\int\cos^5\text{x}\text{ dx}$
$=\int\cos^4\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^2\text{ dt}$
$=\int(1+\text{t}^4-2\text{t}^2)\text{dt}$
$=\int\text{dt}+\int\text{t}^4\text{ dt}-2\int\text{t}^2\text{ dt}$
$=\text{t}+\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}+\text{C}$
$=\int\cos^4\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^2\text{ dt}$
$=\int(1+\text{t}^4-2\text{t}^2)\text{dt}$
$=\int\text{dt}+\int\text{t}^4\text{ dt}-2\int\text{t}^2\text{ dt}$
$=\text{t}+\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}+\text{C}$
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 the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$ and is in the direction $\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}.$
→Find the inverse of the following matrices:
$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
→$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is $\frac{28\times9^6}{10^8}$.
→In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}\text{at x}=0$
→$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}\text{at x}=0$
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=\text{t}+\frac{1}{\text{t}},\text{ y}=\text{t}-\frac{1}{\text{t}}$
→$\text{x}=\text{t}+\frac{1}{\text{t}},\text{ y}=\text{t}-\frac{1}{\text{t}}$
Find the distance of the point P(-1, -5, -10) from the point of intersection of the line joining the points A(2, -1, 2) and B(5, 3, 4) with the plane x - y + z = 5.
The corner points of the feasible region determined by the system of linear inequations are as shown below :
Answer each of the following:
i. Let $z = 13x - 15y $ be the objective function. Find the maximum and minimum values of $z$ and also the
corresponding points at which the maximum and minimum values occur.
ii. Let $z = kx + y $ be the objective function. Find $k,$ if the value of $z$ at $A$ is same as the value of $z$ at $B$.
→Answer each of the following:
i. Let $z = 13x - 15y $ be the objective function. Find the maximum and minimum values of $z$ and also the
corresponding points at which the maximum and minimum values occur.
ii. Let $z = kx + y $ be the objective function. Find $k,$ if the value of $z$ at $A$ is same as the value of $z$ at $B$.
If $y \sqrt{1-x^2}=\sin ^{-1} x$, then find $\frac{d y}{d x}$
→If either vector $\vec{a}=\vec{0}\ \text{or}\ \vec{b}=\vec{0},\ \text{then}\ \vec{a}\cdot\vec{b}=0.$ But the converse need not be true. Justify your answer with an example.
→Integrate the function: $\frac{x}{\sqrt{x+4}}, x>0$
→