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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following:$\int e^{-3x}\cos^3xdx$

$\begin{array}{1 1} (A) \large\frac{e^{-3x}}{24}\normalsize[\sin 3x+\cos 3x]+\large\frac{3e^{-3x}}{40}\normalsize[\sin x+\cos x]+c. \\ (B) \large\frac{e^{-3x}}{24}\normalsize[\sin x+\cos x]-\large\frac{3e^{-3x}}{40}\normalsize[\sin 3x+\cos 3x]+c. \\ (C) \large\frac{e^{-3x}}{24}\normalsize[\sin 3x-\cos 3x]+\large\frac{3e^{-3x}}{40}\normalsize[\sin x-\cos x]+c. \\(D) \large\frac{e^{-3x}}{24}\normalsize[\sin x+\cos x]+\large\frac{3e^{-3x}}{40}\normalsize[\sin 3x+\cos 3x]+c. \end{array} $

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1 Answer

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Toolbox:
  • $\cos 3x=4\cos^3x-3\cos x$
  • $\int u dv=uv-\int v du$
  • $\int e^x dx=e^x+c$
Step 1:
Let $I=\int e^{-3x}.\cos^3x dx.$
$\cos3x=4\cos^3x-3\cos x$
$\cos^3x=\large\frac{\cos 3x+3\cos x}{4}$
Hence $I=\int e^{-3x}\big(\large\frac{\cos 3x+3\cos x}{4}\big)\normalsize dx.$
This can be split and written as
$I=\large\frac{1}{4}\normalsize \int e^{-3x}\cos 3x dx+\large\frac{3}{4}\normalsize\int e^{-3x}\cos x dx$
Let $I=\frac{1}{4}I_1+\frac{3}{4}I_2$
Consider $I_1=\int e^{-3x}\cos 3x dx$
This is of the form $\int u dv$
Hence $\int u dv=uv-\int v du$
Let $u=e^{-3x}$.On differentiating with respect to x,we get
$du=-3e^{-3x}dx.$
Let $dv=\cos 3x dx$,On integrating we get,
$v=\frac{1}{3}\sin 3x$
Now substituting for $u,v,du$ and $dv$ we get,
$I_1=\big(e^{-3x}\frac{1}{3}\sin 3x\big)-\int\frac{1}{3}\sin 3x.(-3e^{-3x})dx.$
Step 2:
$I_1=\large\frac{e^{-3x}}{3}\normalsize\sin 3x+\int \sin 3x.e^{-3x}dx.$------(1)
Consider $\int \sin 3x.e^{-3x}dx.$
This is again of the form $\int u dv$
Let us take $u=e^{-3x}$.On differentiating with respect to x we get
$du=-3e^{-3x}.dx$
Let $dv=\sin 3xdx$ and on integrating we get,
$v=\large\frac{-1}{3}\normalsize\cos 3x$
Now substituting for $u,v,du$ and $dv$ we get,
$\int \sin 3xe^{-3x}dx=e^{-3x}\big(\frac{-1}{3}\cos 3x\big)-\int\frac{-1}{3}\cos 3x.(-3e^{-3x})dx.$
$\qquad\qquad=\large\frac{-e^{-3x}\cos 3x}{3}\normalsize-\int e^{-3x}\cos 3x.dx$
But $\int e^{-3x}.\cos 3x dx=I_1$
On substituting this in equ (1) we get,
Hence $I_1=\frac{e^{-3x}\sin 3x}{3}-\frac{e^{-3x}\cos 3x}{3}-I_1$
$2I_1=\large\frac{e^{-3x}}{3}\normalsize[\sin 3x-\cos 3x]$
$I_1=\large\frac{e^{-3x}}{6}\normalsize[\sin 3x-\cos 3x]$-----(2)
Step 3:
Now consider $I_2=\int e^{-3x}.\cos 3x dx.$
This is again of the form $\int u dv$
Let $u=e^{-3x}$,on differentiating we get,
$du=-3e^{-3x}.dx$
Let $dv=\cos xdx$,on integrating we get
$v=\sin x$
Hence $\int u dv=uv-\int v du$
Now substituting for $u,v,du$ and $dv$ we get,
$I_2=e^{-3x}.\sin x-\int \sin x.(-3e^{-3x}.dx)$
$\;\;=e^{-3x}\sin x+3\int \sin x.e^{-3x}dx$
Now consider $\int \sin x.e^{-3x}dx$
Let $u=e^{-3x}$,on differentiating we get,
$du=-3e^{-3x}$
Let $dv=\sin xdx$,on integrating we get,
$v=-\cos x.$
Now substituting for $u,v,du$ and $dv$ we get,
$\Rightarrow (e^{-3x}(-\cos x))-\int -\cos x(-3e^{-3x})dx.$
$\;\;=-e^{-3x}\cos x-\int 3e^{-3x}\cos x dx.$
But $\int e^{-3x}\cos xdx=I_2$
$I_2=-e^{-3x}\cos 3x-9I_2$
$10I_2=-e^{-3x}\cos x+e^{-3x}\sin x$------(3)
Now we know $I=\large\frac{1}{4}I_1+\frac{3}{4}I_2$
Now combining equ (1) and equ (2)
$I=\frac{1}{4}[\frac{e^{-3x}}{6}(\sin 3x-\cos 3x)]+\frac{3}{4}[\frac{e^{-3x}}{10}(\sin x-\cos x)]$
$\;\;=\large\frac{e^{-3x}}{24}\normalsize[\sin 3x-\cos 3x]+\large\frac{3e^{-3x}}{40}\normalsize[\sin x-\cos x]+c.$
answered Apr 23, 2013 by sreemathi.v
 
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