Find \(\int_{π/4}^{π/2} \,2sinx \,sin⁡(cos⁡x) \,dx\).

Category: QuestionsFind \(\int_{π/4}^{π/2} \,2sinx \,sin⁡(cos⁡x) \,dx\).
Editor">Editor Staff asked 11 months ago

Find \(\int_{π/4}^{π/2} \,2sinx \,sin⁡(cos⁡x) \,dx\).
 
(a) 2(1-cos⁡\(\frac{1}{\sqrt{2}}\))
 
(b) (cos⁡\(\frac{1}{\sqrt{2}}\)-cos⁡1)
 
(c) 2(cos⁡\(\frac{1}{\sqrt{2}}\)+1)
 
(d) (cos⁡\(\frac{1}{\sqrt{2}}\)+cos⁡1)
 
This question was posed to me in semester exam.
 
Question is taken from Fundamental Theorem of Calculus-1 in chapter Integrals of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Right option is (a) 2(1-cos⁡\(\frac{1}{\sqrt{2}}\))
 
For explanation I would say: Let \(I=\int_{π/4}^{π/2} \,2sinx \,sin⁡(cos⁡x) \,dx\)
 
F(x)=\(\int 2 \,sin⁡x \,sin⁡(cos⁡x)dx\)
 
Let cos⁡x=t
 
Differentiating w.r.t x, we get
 
sin⁡x dx=dt
 
∴\(\int 2 \,sin⁡x \,sin⁡(cos⁡x)dx=\int 2 \,sin⁡t \,dt=-2 \,cos⁡t\)
 
Replacing t with cos⁡x, we get
 
∴∫ 2 sin⁡x sin⁡(cos⁡x)dx=-2 cos⁡(cos⁡x)
 
By applying the limits, we get
 
\(I=F(\frac{π}{4})-F(\frac{π}{2})=-2 cos⁡(\frac{cos⁡π}{4})+2 cos⁡(\frac{cos⁡π}{2})\)
 
=2(1-cos⁡\(\frac{1}{\sqrt{2}}\))