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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: sinx+cosx1+2sin2xdx\int{\dfrac{\sin x+\cos x}{\sqrt{1+2\sin 2x}}dx}

Explanation

Solution

Hint: To evaluate the value of following integral square the expression sinx+cosx\sin x+\cos x and simplify it using the identity (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab. Further, simplify the expression using the trigonometric identities cos2x+sin2x=1{{\cos }^{2}}x+{{\sin }^{2}}x=1 and sin2x=2sinxcosx\sin 2x=2\sin x\cos x. Simplify the integral and then use the fact that xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}.

Step-by-step answer:
We have to evaluate the value of the integral sinx+cosx1+2sin2xdx\int{\dfrac{\sin x+\cos x}{\sqrt{1+2\sin 2x}}dx}.
To do so, we will first simplify the expression 1+2sin2x\sqrt{1+2\sin 2x}.
We will firstly square the expression sinx+cosx\sin x+\cos x. We know the algebraic identity (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.
Substituting a=cosx,b=sinxa=\cos x,b=\sin x in the above equation, we have (sinx+cosx)2=sin2x+cos2x+2sinxcosx{{\left( \sin x+\cos x \right)}^{2}}={{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x.
We know the trigonometric identities cos2x+sin2x=1{{\cos }^{2}}x+{{\sin }^{2}}x=1 and sin2x=2sinxcosx\sin 2x=2\sin x\cos x.
Thus, we can rewrite the above equation as (sinx+cosx)2=sin2x+cos2x+2sinxcosx=1+sin2x.....(1){{\left( \sin x+\cos x \right)}^{2}}={{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+\sin 2x.....\left( 1 \right).
Substituting equation (1) in the integral sinx+cosx1+2sin2xdx\int{\dfrac{\sin x+\cos x}{\sqrt{1+2\sin 2x}}dx}, we have sinx+cosx1+2sin2xdx=sinx+cosx(sinx+cosx)2dx\int{\dfrac{\sin x+\cos x}{\sqrt{1+2\sin 2x}}dx}=\int{\dfrac{\sin x+\cos x}{\sqrt{{{\left( \sin x+\cos x \right)}^{2}}}}dx}.
Simplifying the above equation, we have sinx+cosx1+2sin2xdx=sinx+cosx(sinx+cosx)2dx=sinx+cosxsinx+cosxdx=1dx\int{\dfrac{\sin x+\cos x}{\sqrt{1+2\sin 2x}}dx}=\int{\dfrac{\sin x+\cos x}{\sqrt{{{\left( \sin x+\cos x \right)}^{2}}}}dx}=\int{\dfrac{\sin x+\cos x}{\sin x+\cos x}dx}=\int{1dx}.
We know that the integral of xn{{x}^{n}} is xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}.
Substituting n=0n=0 in the above expression, we have x0dx=1dx=x1=x\int{{{x}^{0}}dx}=\int{1dx}=\dfrac{x}{1}=x.
Hence, the value of the integral sinx+cosx1+2sin2xdx\int{\dfrac{\sin x+\cos x}{\sqrt{1+2\sin 2x}}dx} is xx.

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.