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Question: Evaluate the following indefinite integral : \[-2\int{\cos 2\theta \sin 2\theta d\theta }\]...

Evaluate the following indefinite integral :
−2∫cos⁡2θsin⁡2θdθ-2\int{\cos 2\theta \sin 2\theta d\theta }

Explanation

Solution

Hint: By using the formula 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta , we can represent −2sin⁡2θcos⁡2θ-2\sin 2\theta \cos 2\theta as −sin⁡4θ-\sin 4\theta . After that, then use the formula, ∫f(ax+b)dx=F(ax+b)a+C\int{f\left( ax+b \right)}dx=\dfrac{F\left( ax+b \right)}{a}+C where F′(x)=f(x)F'\left( x \right)=f\left( x \right) and also ∫sin⁡θdθ=−cos⁡θ+C\int{\sin \theta d\theta =-\cos \theta }+C to do the problem.

Complete step-by-step answer:
In the question, we are asked to integrate the function −2sin⁡2θcos⁡2θ-2\sin 2\theta \cos 2\theta with respect to it θ\theta . First of all, we will understand what the indefinite integral is. Indefinite Integral is also known by the names antiderivative, inverse derivative, and primitive function. It is said that an indefinite integral of a function f is a differentiable function F whose derivative is equal to the original function f. This can be stated symbolically as F’ = f. The process of solving antiderivatives is called antidifferentiation (or definite integration) and its opposite operation is called differentiation, which is the operation of finding derivatives. Antiderivatives are related to definite integrals through the fundamental theorem of Calculus. The definite integral of a function over an interval is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval.
For example: Let F(x)=x33F\left( x \right)=\dfrac{{{x}^{3}}}{3}, then f(x)=x2f\left( x \right)={{x}^{2}} as it’s a derivative of F(x)=x33F\left( x \right)=\dfrac{{{x}^{3}}}{3}. As the derivative of constant is 0, so x2{{x}^{2}} will have an infinite number of antiderivatives such as x33,x33+1,x33−2, etc.\dfrac{{{x}^{3}}}{3},\dfrac{{{x}^{3}}}{3}+1,\dfrac{{{x}^{3}}}{3}-2,\text{ etc}.Thus F(x) can be represented as x33+C\dfrac{{{x}^{3}}}{3}+C where C is any constant.
So, we have to integrate the function −2sin⁡2θcos⁡2θ-2\sin 2\theta \cos 2\theta . At first, we will use the identity sin 2x = 2 sin x cos x.
So, 2sin⁡2θcos⁡2θ2\sin 2\theta \cos 2\theta can be written as sin⁡(2×2θ)=sin⁡(4θ)\sin \left( 2\times 2\theta \right)=\sin \left( 4\theta \right). Hence, −2sin⁡2θcos⁡2θ-2\sin 2\theta \cos 2\theta can be written as −sin⁡(4θ)-\sin \left( 4\theta \right). Now, we will integrate −sin⁡4θ-\sin 4\theta . So,
−2∫cos⁡2θsin⁡2θdθ=−∫sin⁡4θdθ-2\int{\cos 2\theta \sin 2\theta d\theta =-\int{\sin 4\theta d\theta }}
Now, we will apply the formula which is
∫f(ax+b)dx=F(ax+b)a+C\int{f\left( ax+b \right)}dx=\dfrac{F\left( ax+b \right)}{a}+C
Here, f(ax + b) is a function of (ax + b) where a, b are constants and F’(x) = f(x). Also, we will apply the rule,
∫sin⁡θdθ=−cos⁡θ+C\int{\sin \theta d\theta =-\cos \theta +C}
So, ⇒−∫sin⁡4θdθ=−(−cos⁡4θ)4+C\Rightarrow -\int{\sin 4\theta d\theta =-\dfrac{\left( -\cos 4\theta \right)}{4}+C}
=cos⁡4θ4+C=\dfrac{\cos 4\theta }{4}+C
Hence, the antiderivative of the function −2sin⁡2θcos⁡2θ-2\sin 2\theta \cos 2\theta is cos⁡4θ4+C\dfrac{\cos 4\theta }{4}+C where ‘C’ is any constant value.
Hence, cos⁡4θ4+C\dfrac{\cos 4\theta }{4}+C is the answer.

Note: Students should know the formula or identities related to integration so that they can use it to solve the problem. They should also be careful while calculating integrals to avoid any errors. Be careful about the formulas of integral and derivatives.