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Evaluate $\large \int \frac{2\sin 2 x -\cos x }{6-\cos^2x -4\sin x }$$dx$.

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  • $I=\int f(x) dx$
  • Put $x=g(t)$ then $dx=g'(t)$
  • Therefore $I=\int f(g(t))g'(t)dt$
  • $\int \large\frac{dx}{x^+a^2}=\frac{1}{a}$$\tan^{-1}\big(\Large\frac{x}{a}\big)$
  • $\int \large\frac{dx}{x}$$=\log \mid x\mid+c$
Step 1:
Let $I = \int \large\frac{ (2\sin2x - \cos x)}{(6-\cos^2x-4\sin x)}$$dx$
We know that $\sin 2x=2\sin x\cos x$
$\qquad=\int\large\frac {[4\sin x\cos x - \cos x] dx}{6-1+\sin^2x-4\sin x}$
$\qquad=\int \large\frac{ [\cos x(4\sin x-1)]dx}{[\sin^2x-4\sin x+5]}$
Step 2:
Put $\sin x = t$ then $\cos xdx = dt$
substituting this we get,
$\qquad=\int \large\frac{ [4t-1]dt}{t^2-4t+5}$
$\qquad=\int \large\frac{[4t-8+7]dt}{t^2-4t+5}$
$\qquad=\int 2 \large\frac{[2t-4]dt}{[t^2-4t+5]} +\frac{7dt}{[t^2-4t+5]}$
Step 3:
Put $t^2-4t+5= y$, then $[2t-4]dt = dy$
Therefore $\int 2 \large\frac{ dy}{y }$$+7 \int \large\frac{dt}{(t-2)^2+1}$
Step 4:
On integrating we get,
$2 \log |y| + 7 \tan^{-1}(t-2)$$+C$
Step 5:
Substituting for $y$ and $t$ we get
$2 \log|\sin^2x-4\sin x+5| + 7 \tan^{-1}(\sin x-2)$$+c$
answered Dec 22, 2013 by balaji.thirumalai

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