Question

Integrate

$\displaystyle \int \mathrm{ \bigg[sin (ax+b) cos(ax+b) \bigg] dx.}$


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Answer 1

Given :-

${\quad \leadsto \quad \displaystyle \int \sf \bigg\{ \sin ( ax + b ) \cos ( ax + b ) \bigg\} dx }$

To Find :-

Value of the given indefinite integral

Solution :-

Consider ;

${\quad \leadsto \quad \bf I = \displaystyle \int \sf \bigg\{ \sin ( ax + b ) \cos ( ax + b ) \bigg\} dx}$

From Integration By Parts we knows that :-

${ \bigstar { \underline { \boxed { \red { { \displaystyle \int \bf ( u \cdot v ) dx = \bf \displaystyle \bf u \int v \: dx - \int \bigg\{ \dfrac{du}{dx} \: \int v \: dx \bigg\} dx }}}}}}{\bigstar}$

Now , Integrating by parts we have

${ : \implies \quad \bf I = \displaystyle \sf \sin ( ax + b ) \int \cos ( ax + b ) \: dx - \int \bigg[ \dfrac{d}{dx} \{ \sin ( ax + b ) \} \: \int \cos ( ax + b ) \: dx \bigg] dx}$

We knows that ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { { \bf \dfrac{d}{dx} \{ \sin x \} = \cos x }}}}}}{\bigstar}$

Using this we have ;

${ : \implies \quad \bf I = \displaystyle \sf \sin ( ax + b ) \int \cos ( ax + b ) \: dx - \int \bigg[ \cos ( ax + b ) \times \dfrac{d}{dx} ( ax + b ) \times \: \int \cos ( ax + b ) \: dx \bigg] dx}$

${ : \implies \quad \bf I = \displaystyle \sf \sin ( ax + b ) \int \cos ( ax + b ) \: dx - \int \bigg[ \cos ( ax + b ) \times \bigg\{ \dfrac{d}{dx} ( ax ) + \dfrac{d}{dx} ( b ) \bigg\} \times \: \int \cos ( ax + b ) \: dx \bigg] dx}$

Now , as Differentiation of a constant is 0 , so ;

${ : \implies \quad \bf I = \displaystyle \sf \sin ( ax + b ) \int \cos ( ax + b ) \: dx - \int \bigg[ \cos ( ax + b ) \times \bigg\{ a \times \dfrac{d}{dx} ( x ) + 0 \bigg\} \: \int \cos ( ax + b ) \: dx \bigg] dx}$

${ : \implies \quad \bf I = \displaystyle \sf \sin ( ax + b ) \int \cos ( ax + b ) \: dx - \int \bigg[ a \cos ( ax + b ) \: \int \cos ( ax + b ) \: dx \bigg] dx}$

We knows that ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { { \bf \displaystyle \int \bf \cos ( ax + b ) dx = \dfrac{\sin ( ax + b )}{a} }}}}}}{\bigstar}$

Using this we have ;

${ : \implies \quad \bf I = \displaystyle \sf \sin ( ax + b ) \times \dfrac{\sin ( ax + b)}{a} - \int \bigg[ a \times \cos ( ax + b ) \times \dfrac{\sin ( ax + b)}{a} \bigg] dx}$

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{\sin² ( ax + b)}{a} - \int \bigg[ \cancel{a} \times \cos ( ax + b ) \times \dfrac{\sin ( ax + b)}{\cancel{a}} \bigg] dx}$

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{\sin² ( ax + b)}{a} -\displaystyle \int \sf \bigg\{ \sin ( ax + b ) \cos ( ax + b ) \bigg\} dx}$

This can be written as ;

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{\sin² ( ax + b)}{a} -\bf I}$

${ : \implies \quad \bf I + I= \displaystyle \sf \dfrac{\sin² ( ax + b)}{a} }$

${ : \implies \quad \bf 2I = \displaystyle \sf \dfrac{\sin² ( ax + b)}{a} }$

${ : \implies \quad \bf I = \displaystyle \sf \dfrac{\sin² ( ax + b)}{2a} + C }$

${ \bigstar { \underline { \boxed { \red { { \therefore \bf \displaystyle \int \bf \bigg\{ \sin ( ax + b ) \cos ( ax + b ) \bigg\} dx =\bf \dfrac{\sin² ( ax + b)}{2a} + C }}}}}}{\bigstar}$