$$ \int \frac{e^{\sin x}(\sin 2 x-8 \cos x)}{2(\sin x-3)^2} d x= $$
$e^{\sin x}(\sin x-3)+C$
$\frac{e^{\sin x}}{(\sin x-3)^2}+C$
$e^{\sin x}(\sin x-3)^2+C$
$\frac{e^{\sin x}}{\sin x-3}+C$
If $\int\left(3 t^2 \sin \frac{1}{t}-t \cos \frac{1}{t}\right) d t=f(t) \sin \left(\frac{1}{t}\right)+C$ then $f(2)=$
2
-12
8
-16
$$ \int(\log x)^3 x^4 d x= $$
$x^5\left[\frac{1}{5}(\log x)^3-\frac{4}{25}(\log x)^2-\frac{9}{125} \log x-\frac{8}{125}\right]+C$
$x^5\left[\frac{1}{5}(\log x)^3+\frac{3}{25}(\log x)^2-\frac{6}{125} \log x-\frac{6}{125}\right]+C$
$$ \int \frac{\sin 2 x}{\sin ^2 x+3 \cos x-3} d x $$
$2 \log \left|\frac{\cos x-2}{\cos x-1}\right|+C$
$\log \left(\frac{(\cos x-2)^2}{(\cos x-1)^4}\right)+C$
$\log \left(\frac{(\cos x-2)^2}{|\cos x-1|}\right)+C$
$\log \left(\frac{(\cos x-2)^4}{(\cos x-1)^2}\right)+C$
AP EAPCET Papers
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