Mathematics

# Find the integrals of the functions.i) $sin^3 \, x \, cos^3 \, x$ii) $sin \, x \, sin \, 2x \, sin \, 3x$iii) $sin \, 4x \, sin \, 8x$iv) $\dfrac{1 - cos \, x}{1 + cos \, x}$v) $\dfrac{cos \, x}{1 + cos \, x}$

##### SOLUTION
(i) $I=\displaystyle\int \sin^3x\cos^3xdx=\dfrac{1}{8}\displaystyle\int(\sin 2x)^3dx\Rightarrow \dfrac{1}{32}\displaystyle\int (3\sin 3x-4\sin(6x))dx$
$(\sin 3x=3\sin x-4\sin^3x)$
$I=\dfrac{1}{32}\displaystyle\int 3\sin 2xdx-\dfrac{1}{32}4\sin (6x)dx$
$=\dfrac{3}{32.2}\times -(\cos 2x)+\dfrac{4}{6.32}\cos (6x)+C$
$I=\dfrac{\cos (6x)}{48}-\dfrac{3\cos (2x)}{64}+C$.
(ii) $\displaystyle\int \sin x\sin 2x\sin 3xdx$
$\Rightarrow \displaystyle\int\dfrac{1}{2}\left[\cos(3x-x)-\cos (3x+x)\right]\sin 2xdx$
$\Rightarrow \displaystyle\int \dfrac{1}{2}\sin 2x\cos 2xdx-\displaystyle\dfrac{1}{2}\sin 2x\cos 4xdx$
$\Rightarrow \dfrac{1}{4}\displaystyle\int \sin 4xdx-\dfrac{1}{4}\displaystyle\int (\sin (2x+4x)+\sin (2x-4x)dx)$
$\Rightarrow -\dfrac{1}{16}\cos 4x-\dfrac{1}{4}\displaystyle\int \sin 6xdx+\dfrac{1}{4}\displaystyle\int \sin 2xdx$
$I\Rightarrow \dfrac{1}{24}\cos 6x-\dfrac{1}{16}\cos 4x-\dfrac{1}{8}\cos 2x+C$
(iii) $\displaystyle\int \sin 4x\sin 8xdx$
$\Rightarrow \displaystyle\int \dfrac{1}{2}\left[\cos (8x-4x)-\cos (8x+4x)\right]dx$
$\Rightarrow \displaystyle\int \dfrac{1}{2}(\cos 4x-\cos 12x)dx$
$\Rightarrow \dfrac{1}{8}\sin 4x-\dfrac{1}{24}\sin 12x+C$.
(iv) $\displaystyle\int \dfrac{(1-\cos x)}{(1+\cos x)}dx$
$\Rightarrow \displaystyle\int \dfrac{2\sin^2x/_{2}}{2\cos^2 x/_{2}}dx$
$\Rightarrow \displaystyle\int \dfrac{1-\cos^2x/_{2}}{\cos^2x/_{2}}dx$
(iv) $\displaystyle\int \dfrac{(1-\cos x)}{(1+\cos x)}dx$
$\Rightarrow \displaystyle\int \dfrac{2\sin^2x/_{2}][2\cos^2x/_{2}}dx$
$\Rightarrow \displaystyle\int \dfrac{1-\cos^2x/_{2}}{\cos^2x/_{2}}dx$
$\Rightarrow \displaystyle\int \sin^2x/_{2}dx=\displaystyle\int dx$
$\Rightarrow 2\tan x/_{2}-x+C$
(vi) $\displaystyle\int \dfrac{\cos x}{1+\cos x}dx$
$\Rightarrow \displaystyle\int \left(\dfrac{1+\cos x}{cos x+1}-\dfrac{1}{\cos x+1}\right)dx$
$\Rightarrow \displaystyle\int dx-\displaystyle\int \dfrac{1}{1+\cos x}dx$
$\Rightarrow x-\dfrac{1}{2}\displaystyle\int \sec^2x/_{2}dx$
$\Rightarrow x-\dfrac{1}{2}\times 2\tan x/_{2}+C$
$\Rightarrow x-\tan x/_{2}+C$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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