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  • I have a doubt, kindly clarify. - Integral Calculus - JEE Main-10

Find he value of integral  \int \tan 3x\;dx

  • Option 1)

    3\ln \cos 3x + c

  • Option 2)

    3\ln\sec 3x +c

  • Option 3)

    \frac{1}{3}\ln\cos 3x + c

  • Option 4)

    \frac{1}{3}\ln\sec 3x +c

 

Answers (1)

best_answer

As we have learnt,

 

Integrals for Trigonometric functions -

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

\therefore \int sinxdx=-cosx+c

-

 

 \int\tan 3x dx = \int\frac{\sin 3x}{\cos 3x}dx

Put   \cos 3x = t \Rightarrow -3\sin 3xdx = dt \Rightarrow \sin 3x dx = -\frac{dt}{3}

\int \tan 3x dx = \int \frac{-dt}{3t} = -\frac{1}{3}\ln t = -\frac{1}{3}\ln \cos 3x = \frac{1}{3}\ln \sec 3x

 

 


Option 1)

3\ln \cos 3x + c

Option 2)

3\ln\sec 3x +c

Option 3)

\frac{1}{3}\ln\cos 3x + c

Option 4)

\frac{1}{3}\ln\sec 3x +c

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