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Question

Mathematics Question on Integral Calculus

∫1cosx(sinx+2cosx)dx=∫\dfrac{1}{cosx(sinx+2cosx)} dx=

A

ln∣(1−tanx)∣+Cln|(1-tanx)|+C

B

ln∣3+sinx∣+Cln |3+sinx|+C

C

ln∣2+tanx∣+Cln |2+tanx|+C

D

ln∣1+2secx∣+Cln |1+2secx|+C

E

ln∣2−tanx∣+Cln |2-tanx|+C

Answer

ln∣2+tanx∣+Cln |2+tanx|+C

Explanation

Solution

Given that

∫1cosx(sinx+2cosx)dx∫\dfrac{1}{cosx(sinx+2cosx)} dx

take, I=∫1cosx(sinx+2cosx)dxI = ∫\dfrac{1}{cosx(sinx+2cosx)} dx

      $= ∫\dfrac{1}{cosx.sinx+2cos^{2}x)} dx$

     $= ∫\dfrac{sec^{2}x}{tanx+2} dx$        ⇢⇢[by dividing $cos^{2}x$ on both numerator and denominator]

         take $tanx +2 = u$, then derivate both sides w.r.t x we get 

        $sec^{2}x dx= du$

Bye applying this we can write , ∫1udu∫\dfrac{1}{u} du

                                              $= ln|u|$

                                              $= ln|(tanx+2)|+C$  

                                              $= ln|(2+tanx)|+C$