Answer to Question #135368 in Calculus for Nikhil

Question

Integrate sec^xlog(sinx)dx

Answer 1

$\int sec^x[ln (sin x)]dx = \int sec^{x-2} [ln (sin x)] sec^{2} [ln (sin x)] dx$

Now, we apply integration by Parts by the formula;

$\int udv = uv - \int vdu$

$u = sec^{x-2} [ln (sin x)]$

$dv =sec^{2} [ln (sin x)] dx$

$du = (x-2) sec^{x-3} [ln (sin x)] . sec [ln (sin x)] .tan [ln (sin x)]$

$v = tan [ln (sin x)]$

$\therefore$ $\int sec^x[ln (sin x)]dx = sec^{x-2} [ln (sin x)] tan [ln (sin x)] - (x-2) \int sec^{x-2} [ln (sin x)] tan^2 [ln (sin x)] dx$

But, $tan^2x = sec^2x -1$

$\int sec^x [ln (sin x)]dx = sec^{x-2} [ln (sin x)] tan [ln (sin x)] - (x-2) \int sec^x [ln (sin x)]dx + (x-2) \int sec^{x-2} [ln (sin x)] dx$

$\int sec^x [ln (sin x)]dx + (x-2) \int sec^x [ln (sin x)]dx = sec^{x-2} [ln (sin x)] tan [ln (sin x)] - \sout {(x-2) \int sec^x [ln (sin x)]dx} + (x-2) \int sec^{x-2} [ln (sin x)] dx + \sout{(x-2) \int sec^x [ln (sin x)]dx}$

$\frac{\sout{(x-1)} \int sec^x [ln (sin x)]dx}{\sout{(x-1)}} = \frac{ sec^{x-2} [ln (sin x)] tan [ln (sin x)]}{x-1} + \frac{x-2 \int sec^{x-2} [ln (sin x)] dx}{x-1}$

$Therefore,$

$\int sec^x [ln (sin x)]dx = \frac{ sec^{x-2} [ln (sin x)] tan [ln (sin x)]}{x-1} + \frac{x-2 }{x-1} \int sec^{x-2} [ln (sin x)] dx$

$Where, x \not = 1$