1
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

For $k \in(1, \infty), \int \frac{1}{1+k \cos x} d x=$

A

$\frac{2}{\sqrt{1+k^2}} \tan ^{-1}\left(\sqrt{\frac{1-k}{1+k}} \tan \frac{x}{2}\right)+C$

B

$\frac{1}{\sqrt{k^2-1}} \log \left(\frac{\sqrt{k+1}+\sqrt{k-1} \tan \frac{x}{2}}{\sqrt{k+1}-\sqrt{k-1}}\right)+C$

C

$\frac{1}{\sqrt{k^2+1}} \log ^{-1}\left(\frac{\sqrt{k+1}+\sqrt{k-1} \tan \frac{x}{2}}{\sqrt{k+1}-\sqrt{k-1} \tan \frac{x}{2}}\right)+C$

D

$\frac{1}{\sqrt{k^2-1}} \tan ^{-1}\left(\frac{\sqrt{k-1} \cos \frac{x}{2}+\sqrt{k-1} \sin \frac{x}{2}}{\sqrt{k+1} \cos \frac{x}{2}-\sqrt{k-1} \sin \frac{x}{2}}\right)+C$

2
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int e^{-3 x}\left(x^2+\sin 4 x\right) d x= $$

A

$-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C$

B

$-e^{-3 x}\left(\frac{x^2}{3}-\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C$

C

$-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x-\frac{4}{25} \cos 4 x\right)+C$

D

$-e^{-3 x}\left(\frac{x^2}{3}-\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x-\frac{4}{25} \cos 4 x\right)+C$

3
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$, then $\frac{m-l}{r}=$

A

3

B

4

C

5

D

6

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