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Question: Solve: \[\int {\dfrac{{\sin \theta }}{{(4 + {{\cos }^2}\theta )(2 - {{\sin }^2}\theta )}}d\theta } \...

Solve: ∫sin⁡θ(4+cos⁡2θ)(2−sin⁡2θ)dθ\int {\dfrac{{\sin \theta }}{{(4 + {{\cos }^2}\theta )(2 - {{\sin }^2}\theta )}}d\theta }

Explanation

Solution

Integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. Here, we are given the integration and we need to find the value of it. As we know the trigonometry identity which we will use here is cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 . We will assume cosx = t and then find the derivative of it. After solving this, we will substitute the value at the end and thus, will get the final output.

Complete step by step answer:
We know that the trigonometric ratios of a triangle are also called the trigonometric functions. Sine, cosine, and tangent are 3 important trigonometric functions and are abbreviated as sin, cos and tan. And thus, we will use this to solve the problem.Given that,
∫sin⁡θ(4+cos⁡2θ)(2−sin⁡2θ)dθ\int {\dfrac{{\sin \theta }}{{(4 + {{\cos }^2}\theta )(2 - {{\sin }^2}\theta )}}d\theta }
∫sin⁡θ(4+cos⁡2θ)(1+(1−sin⁡2θ))dθ\int {\dfrac{{\sin \theta }}{{(4 + {{\cos }^2}\theta )(1 + (1 - {{\sin }^2}\theta ))}}d\theta }
We know that, 1−sin⁡2θ=cos⁡2θ1 - {\sin ^2}\theta = {\cos ^2}\theta
Substituting this value, we will get,
∫sin⁡θ(4+cos⁡2θ)(1+cos⁡2θ)dθ\int {\dfrac{{\sin \theta }}{{(4 + {{\cos }^2}\theta )(1 + {{\cos }^2}\theta )}}d\theta }

Let x=cos⁡θx = \cos \theta and so dx=−sin⁡θdθdx = - \sin \theta d\theta
Substituting this value, we will get,
−∫dx(4+x2)(1+x2)- \int {\dfrac{{dx}}{{(4 + {x^2})(1 + {x^2})}}}
⇒−13∫(1(1+x2)−1(4+x2))dx\Rightarrow - \dfrac{1}{3}\int {(\dfrac{1}{{(1 + {x^2})}} - \dfrac{1}{{(4 + {x^2})}})dx}
⇒13∫(1(4+x2)−1(1+x2))dx\Rightarrow \dfrac{1}{3}\int {(\dfrac{1}{{(4 + {x^2})}} - \dfrac{1}{{(1 + {x^2})}})dx}
⇒13[12tan⁡−1(x2)−tan⁡−1x)]+c\Rightarrow \dfrac{1}{3}[\dfrac{1}{2}{\tan ^{ - 1}}(\dfrac{x}{2}) - {\tan ^{ - 1}}x)] + c
⇒13[12tan⁡−1(cos⁡θ2)−tan⁡−1(cos⁡θ)]+c\Rightarrow \dfrac{1}{3}[\dfrac{1}{2}{\tan ^{ - 1}}(\dfrac{{\cos \theta }}{2}) - {\tan ^{ - 1}}(\cos \theta )] + c
⇒16tan⁡−1(cos⁡θ2)−13tan⁡−1(cos⁡θ)+c\Rightarrow \dfrac{1}{6}{\tan ^{ - 1}}(\dfrac{{\cos \theta }}{2}) - \dfrac{1}{3}{\tan ^{ - 1}}(\cos \theta ) + c

Hence, the value of the given integral is ∫sin⁡θ(4+cos⁡2θ)(2−sin⁡2θ)dθ=16tan⁡−1(cos⁡θ2)−13tan⁡−1(cos⁡θ)+c\int {\dfrac{{\sin \theta }}{{(4 + {{\cos }^2}\theta )(2 - {{\sin }^2}\theta )}}d\theta } = \dfrac{1}{6}{\tan ^{ - 1}}(\dfrac{{\cos \theta }}{2}) - \dfrac{1}{3}{\tan ^{ - 1}}(\cos \theta ) + c.

Note: Integration is the calculation of an integral. In short, the integration denotes the summation of discrete data. Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc., that occurs due to a collection of small data, which cannot be measured singularly. We know that there are two major types of calculus:- 1) Differential Calculus and 2) Integral Calculus. Trigonometry is one of those divisions in mathematics that helps in finding the angles and missing sides of a triangle with the help of trigonometric ratios. The angles are either measured in radians or degrees.