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Question

Question: What is the integral of \(\operatorname{Sin}2x\)?...

What is the integral of Sin2x\operatorname{Sin}2x?

Explanation

Solution

To integrate the trigonometric function we will use the basic concept of integrating the simple trigonometric function, likesin(ax+b)\sin \left( ax+b \right)where ‘a’ and ‘b’ are constant and ‘x’ is variable.

Complete step-by-step solution:
Moving ahead with the question in step wise manner,
We are asked to integrateSin2x\operatorname{Sin}2x. Since we directly know the integration ofsinx\sin x which iscosx-\cos x. But integration ofSin2x\operatorname{Sin}2x is a little bit complex. For this type of question we know that we can integrate the trigonometric function like integration ofsin(ax+b)\sin \left( ax+b \right) (where ‘a’ and ‘b’ are constant and ‘x’ is variable) which is equalcos(ax+b)a+c\dfrac{-\cos \left( ax+b \right)}{a}+c, i.e.sin(ax+b)=cos(ax+b)a+c\int{\sin \left( ax+b \right)}=\dfrac{-\cos \left( ax+b \right)}{a}+c .
So using the same formula let us integrateSin2x\operatorname{Sin}2x. By comparing it with the formula we can say it is written asSin(2x+0)\operatorname{Sin}\left( 2x+0 \right)in which constant ‘a’ and ‘b’ are ‘2’ and ‘0’ respectively.
So according to question we had to find out the integration ofSin(2x+0)\operatorname{Sin}\left( 2x+0 \right)which we can write it asSin(2x+0)\int{\operatorname{Sin}\left( 2x+0 \right)}.
As by formula we know that integration of trigonometric functionsinx\sin xiscosx-\cos xso by using the formula ofsin(ax+b)\sin \left( ax+b \right)we will get;
sin2x=Sin(2x+0) sin2x=cos(2x+0)2+c sin2x=cos(2x)2+c \begin{aligned} & \int{\sin 2x=}\int{\operatorname{Sin}\left( 2x+0 \right)} \\\ & \int{\sin 2x=}\dfrac{-\cos \left( 2x+0 \right)}{2}+c \\\ & \int{\sin 2x=}\dfrac{-\cos \left( 2x \right)}{2}+c \\\ \end{aligned}
So we gotcos(2x+0)2+c\dfrac{-\cos \left( 2x+0 \right)}{2}+cin which c is some constant.
Hence the answer iscos(2x+0)2+c\dfrac{-\cos \left( 2x+0 \right)}{2}+c.

Note: If the angle inside the trigonometric function is present in linear form then only this formula as in our case it is2x2xand if it will be2x22{{x}^{2}}orx3{{x}^{3}}or some other exponential then we can’t apply this method.