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Question: Evaluate the integral \(\int {{{\sin }^2}} \left( {2x + 5} \right)dx\)....

Evaluate the integral ∫sin⁡2(2x+5)dx\int {{{\sin }^2}} \left( {2x + 5} \right)dx.

Explanation

Solution

Hint : To evaluate the required integral, first we will use the trigonometric formula which is given by sin⁡2θ=1−cos⁡2θ2{\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}. After that we will use the sum rule of integration. Then, we will use the integration formula to integrate the function cos⁡(aθ+b)\cos \left( {a\theta + b} \right).

Complete step-by-step answer :
In the given problem, we have to evaluate the integral ∫sin⁡2(2x+5)dx\int {{{\sin }^2}} \left( {2x + 5} \right)dx. Let us say I=∫sin⁡2(2x+5)dxI = \int {{{\sin }^2}\left( {2x + 5} \right)dx} . Now we are going to use the trigonometric formula which is given by sin⁡2θ=1−cos⁡2θ2{\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}. So, we can write
I=∫[1−cos⁡2(2x+5)2]dxI = \int {\left[ {\dfrac{{1 - \cos 2\left( {2x + 5} \right)}}{2}} \right]dx}
In the above integral, let us use the property of integral which is given by ∫kf(x)dx=k∫f(x)dx\int {kf\left( x \right)dx = k\int {f\left( x \right)dx} } . So, we can write
I=12∫[1−cos⁡(4x+10)]dxI = \dfrac{1}{2}\int {\left[ {1 - \cos \left( {4x + 10} \right)} \right]dx}
In above integral, we will use the sum rule of integration which is given by ∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \int {f\left( x \right)dx \pm \int {g\left( x \right)dx} } } . So, we can write
I=12[∫1 dx−∫cos⁡(4x+10)dx]I = \dfrac{1}{2}\left[ {\int {1\,dx - \int {\cos \left( {4x + 10} \right)dx} } } \right]
Now we are going to use the integration formula which is given by ∫cos⁡(ax+b)dx=sin⁡(ax+b)a\int {\cos \left( {ax + b} \right)dx = \dfrac{{\sin \left( {ax + b} \right)}}{a}} . Also note that ∫1 dx=x\int {1\,dx = x} . After integration, we will include a constant which is called integrating constant. So, we can write
I=12[x−sin⁡(4x+10)4]+cI = \dfrac{1}{2}\left[ {x - \dfrac{{\sin \left( {4x + 10} \right)}}{4}} \right] + c
Let us simplify the above expression. So, we can write
I=x2−sin⁡(4x+10)8+cI = \dfrac{x}{2} - \dfrac{{\sin \left( {4x + 10} \right)}}{8} + c
Hence, the required integral is x2−sin⁡(4x+10)8+c\dfrac{x}{2} - \dfrac{{\sin \left( {4x + 10} \right)}}{8} + c. Hence, we can say that ∫sin⁡2(2x+5)dx=x2−sin⁡(4x+10)8+c\int {{{\sin }^2}} \left( {2x + 5} \right)dx = \dfrac{x}{2} - \dfrac{{\sin \left( {4x + 10} \right)}}{8} + c where cc is integrating constant.

Note : In this type of problems of integration, we must remember the trigonometric identities and formulas. Also remember the formula of integration. There is one useful trigonometric formula which is given by cos⁡2θ=1+cos⁡2θ2{\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}.