# How do you integrate sin(ln(4x+5))?

May 27, 2018

$I = \frac{4 x + 5}{8} \left[\sin \left(\ln \left(4 x + 5\right)\right) - \cos \left(\ln \left(4 x + 5\right)\right)\right] + C$

#### Explanation:

Here,

$I = \int \sin \left(\ln \left(4 x + 5\right)\right) \mathrm{dx}$

Subst. color(violet)(ln(4x+5)=u=>4x+5=e^u

$\implies 4 \mathrm{dx} = {e}^{u} \mathrm{du} \implies \mathrm{dx} = \frac{1}{4} {e}^{u} \mathrm{du}$

So,

$I = \int \sin u \times \frac{1}{4} {e}^{u} \mathrm{du}$

$\implies 4 I = \int \sin u {e}^{u} \mathrm{du} \ldots \to \left(A\right)$

$\text{Using "color(blue)"Integration by Parts :}$

color(red)(intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx

$\implies 4 I = \sin u \int {e}^{u} \mathrm{du} - \int \left(\cos u \int {e}^{u} \mathrm{du}\right) \mathrm{du}$

$\implies 4 I = \sin u \cdot {e}^{u} - \int \cos u {e}^{u} \mathrm{du}$

$\implies 4 I = \sin u {e}^{u} - {I}_{1.} . . \to \left(B\right)$

Where, ${I}_{1} = \int \cos u {e}^{u} \mathrm{du}$

Again $\text{using "color(blue)"Integration by Parts :}$

$\therefore {I}_{1} = \cos u \int {e}^{u} \mathrm{du} - \int \left(- \sin u \int {e}^{u} \mathrm{du}\right) \mathrm{du}$

$\therefore {I}_{1} = \cos u {e}^{u} + \int \sin u {e}^{u} \mathrm{du} + c '$

$\therefore {I}_{1} = \cos u {e}^{u} + 4 I + c ' \ldots \to$ from $\left(A\right)$

From $\left(B\right)$ we get,

$\therefore 4 I = \sin u {e}^{u} - \left\{\cos u {e}^{u} + 4 I\right\} + c , w h e r e , c = - c '$

$\therefore 4 I = \sin u {e}^{u} - \cos u {e}^{u} - 4 I + c$

$4 I + 4 I = \sin u {e}^{u} - \cos u {e}^{u} + c$

$8 I = {e}^{u} \left(\sin u - \cos u\right) + c$

$I = \frac{1}{8} {e}^{u} \left(\sin u - \cos u\right) + C , w h e r e , C = \frac{c}{8}$

Subst, back , color(violet)(u=ln(4x+5) and e^u=4x+5

$I = \frac{1}{8} \left(4 x + 5\right) \left[\sin \left(\ln \left(4 x + 5\right)\right) - \cos \left(\ln \left(4 x + 5\right)\right)\right] + C$

$I = \frac{4 x + 5}{8} \left[\sin \left(\ln \left(4 x + 5\right)\right) - \cos \left(\ln \left(4 x + 5\right)\right)\right] + C$