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\int \frac{dx}{cosx-sinx}    is equal to

  • Option 1)

    \frac{1}{\sqrt{2}}log\left | tan\left ( \frac{x}{2}-\frac{3\pi }{8} \right ) \right |+C

  • Option 2)

    \frac{1}{\sqrt{2}}log\left | cot\left ( \frac{x}{2} \right ) \right |+C

  • Option 3)

    \frac{1}{\sqrt{2}}log\left | tan\left ( \frac{x}{2}-\frac{\pi }{8} \right ) \right |+C

  • Option 4)

    \frac{1}{\sqrt{2}}log\left | tan\left ( \frac{x}{2}+\frac{3\pi }{8} \right ) \right |+C

 

Answers (1)

As learnt in concept

Integrals for Trigonometric functions -

\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos x \right ) =sinx

\therefore \int sinxdx=-cosx+c

-

 

 cosx-sinx=\sqrt{2}(\frac{1}{\sqrt{2}}cosx-\frac{1}{\sqrt{2}}sinx)

=\sqrt{2}cos(x+\frac{\pi }{4})

\int \frac{dx}{cosx-sinx}=\frac{1}{\sqrt{2}}\int \frac{dx}{cos(x+\frac{\pi }{4})}=\frac{1}{\sqrt{2}}\int sec(x+\frac{\pi }{4}) dx

=\frac{1}{\sqrt{2}}log\left | tan(\frac{x}{2}+\frac{3\pi }{8}) \right |+C


Option 1)

\frac{1}{\sqrt{2}}log\left | tan\left ( \frac{x}{2}-\frac{3\pi }{8} \right ) \right |+C

This option is incorrect

Option 2)

\frac{1}{\sqrt{2}}log\left | cot\left ( \frac{x}{2} \right ) \right |+C

This option is incorrect

Option 3)

\frac{1}{\sqrt{2}}log\left | tan\left ( \frac{x}{2}-\frac{\pi }{8} \right ) \right |+C

This option is incorrect

Option 4)

\frac{1}{\sqrt{2}}log\left | tan\left ( \frac{x}{2}+\frac{3\pi }{8} \right ) \right |+C

This option is correct

Posted by

Sabhrant Ambastha

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