∫dxa2sin2⁡x+b2cos2⁡x is equal to

# $\int \frac{\mathrm{dx}}{{\mathrm{a}}^{2}{\mathrm{sin}}^{2}\mathrm{x}+{\mathrm{b}}^{2}{\mathrm{cos}}^{2}\mathrm{x}}$ is equal to

1. A

$\frac{1}{\mathrm{ab}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{atan}\mathrm{x}}{\mathrm{b}}\right)+\mathrm{C}$

2. B

$\frac{\mathrm{a}}{\mathrm{b}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{atan}\mathrm{x}}{\mathrm{b}}\right)+\mathrm{C}$

3. C

$\frac{\mathrm{a}}{\mathrm{b}}{\mathrm{tan}}^{-1}\left(\frac{\mathrm{atan}\mathrm{x}}{\mathrm{b}}\right)-\mathrm{C}$

4. D

None of the above

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### Solution:

$\int \frac{\mathrm{dx}}{{\mathrm{a}}^{2}{\mathrm{sin}}^{2}\mathrm{x}+{\mathrm{b}}^{2}{\mathrm{cos}}^{2}\mathrm{x}}=\int \frac{{\mathrm{sec}}^{2}\mathrm{xdx}}{{\mathrm{a}}^{2}{\mathrm{tan}}^{2}\mathrm{x}+{\mathrm{b}}^{2}}=\frac{1}{{\mathrm{a}}^{2}}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{2}+{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{2}}$

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