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Question: Find the value of \(\int {\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx} \)...

Find the value of ∫sin⁡(ax+b)cos⁡(ax+b)dx\int {\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx}

Explanation

Solution

Hint : sine and cosine are the functions of trigonometry. The differentiation of sine is cosine and differentiation of cosine is negative sine. Here in the given integral, consider sin⁡(ax+b)\sin \left( {ax + b} \right) as some variable and find its differentiation and substitute that variable in the place of sin⁡(ax+b)\sin \left( {ax + b} \right) in the question.

Complete step-by-step answer :
We are given to find the value of the integral ∫sin⁡(ax+b)cos⁡(ax+b)dx\int {\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx}
Here, as we can see there is a sine function and a cosine function in the integral with the same angle.
Consider sin⁡(ax+b)\sin \left( {ax + b} \right) as t.
sin⁡(ax+b)=t\sin \left( {ax + b} \right) = t
Now differentiate the above equation with respect to x.
sin⁡(ax+b)=t ddxsin⁡(ax+b)=dtdx→eq(1)  \sin \left( {ax + b} \right) = t \\\ \dfrac{d}{{dx}}\sin \left( {ax + b} \right) = \dfrac{{dt}}{{dx}} \to eq(1) \\\
Differentiation of \sin \left( {ax + b} \right)\] is $$a.\cos \left( {ax + b} \right) and differentiation of a constant is zero. Here b is the constant.
Substitute the value of differentiation of sin⁡(ax+b)\sin \left( {ax + b} \right) in eq (1).
a×cos⁡(ax+b)=dtdxa \times \cos \left( {ax + b} \right) = \dfrac{{dt}}{{dx}}
Send the dx to the LHS and ‘a’ to the RHS
cos⁡(ax+b)dx=dta\cos \left( {ax + b} \right)dx = \dfrac{{dt}}{a}
Substitute the value t in the place of sin⁡(ax+b)\sin \left( {ax + b} \right) and the value dta\dfrac{{dt}}{a} in the place of cos⁡(ax+b)dx\cos \left( {ax + b} \right)dx
So the final integral after substituting values will be
∫t×dta\int {t \times \dfrac{{dt}}{a}}
‘a’ is a constant so take it out.
=1a∫tdt =1a×t22+c (∵∫xndx=xn+1n+1) =t22a+c  = \dfrac{1}{a}\int {tdt} \\\ = \dfrac{1}{a} \times \dfrac{{{t^2}}}{2} + c \\\ \left( {\because \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right) \\\ = \dfrac{{{t^2}}}{{2a}} + c \\\
Substitute the value of t as [\sin \left( {ax + b} \right)$$ in the above step.
=(sin⁡(ax+b))22a+c =sin⁡2(ax+b)2a+c  = \dfrac{{{{\left( {\sin \left( {ax + b} \right)} \right)}^2}}}{{2a}} + c \\\ = \dfrac{{{{\sin }^2}\left( {ax + b} \right)}}{{2a}} + c \\\
Therefore, the value of integral ∫sin⁡(ax+b)cos⁡(ax+b)dx\int {\sin \left( {ax + b} \right)\cos \left( {ax + b} \right)dx} is sin⁡2(ax+b)2a+c\dfrac{{{{\sin }^2}\left( {ax + b} \right)}}{{2a}} + c

Note : Sine and Cosine are the functions of trigonometry revealing the shape of a right triangle. Sine is the ratio of the opposite side of the angle and the hypotenuse whereas Cosine is the ratio of adjacent side of the angle and the hypotenuse. The sum of squares of sine and cosine is always 1. Sine and Cosine are also known as co-functions because sine of an angle is the complement of cosine of the same angle.