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Question: Find the value of \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{1000}}xdx}}{{{{\sin }^{1000}}...

Find the value of ∫0π2sin⁡1000xdxsin⁡1000x+cos⁡1000x\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{1000}}xdx}}{{{{\sin }^{1000}}x + {{\cos }^{1000}}x}}} is equal to
A) 1000
B) 1
C)π2\dfrac{\pi }{2}
D)π4\dfrac{\pi }{4}

Explanation

Solution

Hint: Here we will solve the problem by using the definite integral property ∫abf(x)dx=∫abf(a+b−x)dx\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} }

Complete step-by-step answer:
Given value is ∫0π2sin⁡1000xdxsin⁡1000x+cos⁡1000x\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{1000}}xdx}}{{{{\sin }^{1000}}x + {{\cos }^{1000}}x}}}
Now by using the definite integral property i.e. ∫abf(x)dx=∫abf(a+b−x)dx\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} }
Let us solve the problem
Now if we consider each term in the numerator and denominator of given value as f(x)f(x)
Then we write the value as
I = ∫0π2sin⁡1000(π2−x)sin⁡1000(π2−x)+cos⁡1000(π2−x)dx→1\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{1000}}(\dfrac{\pi }{2} - x)}}{{{{\sin }^{1000}}(\dfrac{\pi }{2} - x) + {{\cos }^{1000}}(\dfrac{\pi }{2} - x)}}} dx \to 1

We know that
sin⁡(π2−x)=cos⁡x\sin (\dfrac{\pi }{2} - x) = \cos x ⇒sin⁡1000(π2−x)=cos⁡1000x \Rightarrow {\sin ^{1000}}(\dfrac{\pi }{2} - x) = {\cos ^{1000}}x
cos⁡(π2−x)=sin⁡x⇒cos⁡1000(π2−x)=sin⁡x\cos (\dfrac{\pi }{2} - x) = \sin x \Rightarrow {\cos ^{1000}}(\dfrac{\pi }{2} - x) = \sin x
From this we can rewrite the value as
⇒∫0π2cos⁡1000xdxcos⁡1000x+sin⁡1000x→2\Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^{1000}}xdx}}{{{{\cos }^{1000}}x + {{\sin }^{1000}}x}}} \to 2
Now by adding equation 1 and 2 we get the value as
2I=∫0π2cos⁡1000x+sin⁡1000xcos⁡1000x+sin⁡1000xdx2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^{1000}}x + {{\sin }^{1000}}x}}{{{{\cos }^{1000}}x + {{\sin }^{1000}}x}}} dx
Here in the above term, numerator and denominator has same value so it get cancels and the value after cancellation is 1
So the term can be written as
⇒2I=∫0π21dx\Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}}{1dx}
We know that ∫1dx=x\int {1dx = x} so let us apply the limits for xx term
⇒2I⇒[x]0π2\Rightarrow 2I \Rightarrow {[x]_0}^{\dfrac{\pi }{2}}
⇒2I=π2−0\Rightarrow 2I = \dfrac{\pi }{2} - 0
⇒I=π4\Rightarrow I = \dfrac{\pi }{4}
Therefore the given value is equals to π4\dfrac{\pi }{4}
Option D is the correct

Note: Make a note that we have to apply definite integral properties for this kind of problem. If needed conversions of values have to be done like the above solution.