Grade 12Integral Calculus

evaluate the following integral
∫ 1 / cos 3x – cos x dx

taniska

11 Years agoGrade 12

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello student,
Please find the answer to your question below

$I = \int \frac{1}{cos3x-cosx}dx$
$cos3x - cosx = 2sin(2x)sin(-x)$
$cos3x - cosx = -2sin(2x)sin(x)$
$I = \int \frac{-1}{2sin2xsinx}dx$
$I = \int \frac{-1}{2.2sinxcosxsinx}dx$
$I = \int \frac{-1}{4sin^{2}xcosx}dx$
$I = \int \frac{-cosx}{4sin^{2}xcos^{2}x}dx$
$sinx = t$
$cosxdx = dt$
$I = \int \frac{-1}{4t^{2}(1-t^{2})}dt$
$I = -\int \frac{1-t^{2}+t^{2}}{4t^{2}(1-t^{2})}dt$
$I = -\int \frac{1}{4t^2}dt-\int\frac{1}{4(1-t^{2})}dt$
$I = \frac{1}{4t}-\frac{1}{4}log(\frac{1+t}{1-t})+c$
$I = \frac{1}{4sinx}-\frac{1}{4}log(\frac{1+sinx}{1-sinx})+c$