Question

please help me solve the sum given in the image

Answer 1

Hey there !!!

$\int\ sinx+cosx/9+16sin2x\, dx$

Let, sinx-cosx=t------------Equation 1

Squaring equation 1

   (sinx-cosx)²=t²

sin²x+cos²x-2sinxcosx=t²

1-sin2x=t²

1-t²=sin2x

Now,

sinx-cosx=t

differentiating wrt to t

(sinx+cosx)dx=dt  , sinx-cosx=t , sin2x=1-t²

$\int\ sinx+cosx/9+16sin2x\, dx$

$\int\1/9+16(1-t^2)\, dt$

$\int\1/25-16t^2\, dt$

$\int\1/5^2-(4t)^2 \, dt$------Equation 2


Equation 2 is of the form :
$\int\1/a^2-x^2\, dx$ =$log(a+x)/2a*log(a-x)$

$\int\1/5^2-(4t)^2 \, dt$ = 
=$4log(5+4t)/2*5log(5-4t)$


$= 2log(5+4(sinx-cosx))/5log(5-4(sinx-cosx)$