Question
Question: Evaluate the value of the following integral: \(\int{\dfrac{\sin x+\cos x}{\sqrt{1+2\sin 2x}}dx}\)...
Evaluate the value of the following integral: ∫1+2sin2xsinx+cosxdx
Solution
Hint: To evaluate the value of following integral square the expression sinx+cosx and simplify it using the identity (a+b)2=a2+b2+2ab. Further, simplify the expression using the trigonometric identities cos2x+sin2x=1 and sin2x=2sinxcosx. Simplify the integral and then use the fact that ∫xndx=n+1xn+1.
Step-by-step answer:
We have to evaluate the value of the integral ∫1+2sin2xsinx+cosxdx.
To do so, we will first simplify the expression 1+2sin2x.
We will firstly square the expression sinx+cosx. We know the algebraic identity (a+b)2=a2+b2+2ab.
Substituting a=cosx,b=sinx in the above equation, we have (sinx+cosx)2=sin2x+cos2x+2sinxcosx.
We know the trigonometric identities cos2x+sin2x=1 and sin2x=2sinxcosx.
Thus, we can rewrite the above equation as (sinx+cosx)2=sin2x+cos2x+2sinxcosx=1+sin2x.....(1).
Substituting equation (1) in the integral ∫1+2sin2xsinx+cosxdx, we have ∫1+2sin2xsinx+cosxdx=∫(sinx+cosx)2sinx+cosxdx.
Simplifying the above equation, we have ∫1+2sin2xsinx+cosxdx=∫(sinx+cosx)2sinx+cosxdx=∫sinx+cosxsinx+cosxdx=∫1dx.
We know that the integral of xn is ∫xndx=n+1xn+1.
Substituting n=0 in the above expression, we have ∫x0dx=∫1dx=1x=x.
Hence, the value of the integral ∫1+2sin2xsinx+cosxdx is x.
Note: We observe that this is an indefinite integral. An indefinite integral is a function that takes the antiderivative of another function. It represents a family of functions whose derivatives are the function given in the integral. It’s necessary to use the trigonometric identities to evaluate the value of the given integral. Otherwise; we won’t be able to solve the question.