Integral Calculus> evaluate the following integral ∫ 1 / sin...

evaluate the following integral
∫ 1 / sin x cos^2 (x) dx

taniska , 11 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 11 Years ago

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

$I = \int \frac{1}{sinx.cos^{2}x}dx$
$I = \int \frac{secx}{sinx.cosx}dx$
$I = \int \frac{secx.tanx}{sinx.cosx.tanx}dx$
$I = \int \frac{secx.tanx}{sin^{2}x}dx$
$I = \int \frac{secx.tanx}{1-cos^{2}x}dx$
$secx = t$
$secxtanxdx = dt$
$I = \int \frac{1}{1-\frac{1}{t^2}}dt$
$I = \int \frac{t^2}{t^2-1}dt$
$I = \int \frac{t^2-1+1}{t^2-1}dt$
$I = \int dt + \int \frac{1}{t^2-1}dt$
$I = t + \int \frac{1}{t^2-1}dt$
$I = t + \frac{1}{2}\int \frac{(t+1)-(t-1)}{(t+1)(t-1)}dt$
$I = t + \frac{1}{2}\int \frac{1}{t-1}dt-\frac{1}{2}\int \frac{1}{t+1}dt$
$I = t + \frac{1}{2}log(\frac{t-1}{t+1})+c$
$I = secx + \frac{1}{2}log(\frac{secx-1}{secx+1})+c$