Mathematics

# $I = \int{{{\cos }^4}x.dx}$

##### SOLUTION
$\displaystyle \int \cos^4 x dx$
Now $\cos^4 x = \cos^3 x . \cos x$
$\displaystyle\int \cos^4 x dx = \int \cos^3 x d (\sin x)$
$\displaystyle\int \cos^4 x dx = \sin x \cos^3 x - \int \sin x d(\cos^3 x) \rightarrow$ By parts
$\displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int sin^2 x \cos^2 x \, dx$
$\displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int (1 - \cos^2 x) \cos^2 x \, dx$
$\displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int \cos^2 x dx - 3 \int \cos^4 x \, dx$
$4 \displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int \cos^2 x \, dx$
$\displaystyle\int \cos^4 x dx = \dfrac{\sin x \, \cos^3 x}{4} + \dfrac{3}{4} \int \cos^2 x dx$
Now, $\displaystyle\int \cos^2 x dx = \int \left(\dfrac{\cos 2x + 1}{2} \right) dx$
$\therefore \displaystyle\int \cos^2 x \, dx = \dfrac{\sin 2x }{4} + \dfrac{x}{2}$
$\therefore \displaystyle\int \cos^4 x \, dx = \dfrac{\sin x \cos^3 x}{4} + \dfrac{3}{4} \left(\dfrac{\sin 2x}{4} + \dfrac{x}{2} \right)$
$= \displaystyle\int \cos^4 x \, dx = \dfrac{\sin x \, \cos^3 x}{4} + \dfrac{3 \, \sin 2x}{16} + \dfrac{3x}{8}$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \int { { e }^{ x } } \left[ \tan { x } -\log { \left( \cos { x } \right) } \right] dx=$
• A. ${ e }^{ x }\log { \left( \ cosec { x } \right) } +c$
• B. ${ e }^{ x }\log { \left( \cos { x } \right) } +c$
• C. ${ e }^{ x }\log { \left( \sin { x } \right) } +c$
• D. ${ e }^{ x }\log { \left( \sec { x } \right) } +c$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
Evaluate:$\displaystyle \int \frac{dx}{\sqrt{15-8x^{2}}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\dfrac{{\sin x + \cos x}}{{\sqrt {1 + \sin 2x} }}dx$ is
• A. $\sin x+c$
• B. $\cos x+c$
• C. $\dfrac{1}{2}(\sin x+\cos x)$
• D. $x+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
lf $f(x)=\displaystyle \frac{x+2}{2x+3}$ , then $\displaystyle \int(\frac{f(x)}{x^{2}})^{1/2}dx$ is

equal to
$\displaystyle \frac{1}{\sqrt{2}}g(\frac{1+\sqrt{2f(x)}}{1-\sqrt{2f(x)}})-\sqrt{\frac{2}{3}}h(\frac{\sqrt{3f(x)}+\sqrt{2}}{\sqrt{3f(x)}-\sqrt{2}})+c$
where
• A. $g(x)=\tan^{-1}x,\ h(x)=\log|x|$
• B. $g(x)=\log|x|,h(x)=\tan^{-1}x$
• C. $g(x)=h(x)=\tan^{-1}x$
• D. $g(x)=\log|x|, h(x)=\log|x|$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$