Definite Integrals
The value of ∫0π/2sin2025x+cos2025xsin2025xdx is:
Select the correct option:
Solution
π/4
Let I=∫0π/2sinnx+cosnxsinnxdx. Using Property ∫0af(x)dx=∫0af(a−x)dx: Replace x with π/2−x. Note that sin(π/2−x)=cosx and cos(π/2−x)=sinx. I=∫0π/2cosnx+sinnxcosnxdx.
Adding the two expressions for I: 2I=∫0π/2sinnx+cosnxsinnx+cosnxdx 2I=∫0π/21dx=[x]0π/2=π/2. So, I=π/4.
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About This Question
- Subject
- mathematics
- Chapter
- integral calculus
- Topic
- definite integrals
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
π/4
Let I=∫0π/2sinnx+cosnxsinnxdx. Using Property ∫0af(x)dx=∫0af(a−x)dx: Replace x with π/2−x. Note that sin(π/2−x)=cosx and cos(π/2−x)=sinx. I=∫0π/2cosnx+sinnxcosnxdx.
Adding the two expressions for I: 2I=∫0π/2sinnx+cosnxsinnx+cosnxdx 2I=∫0π/21dx=[x]0π/2=π/2. So, I=π/4.
This easy difficulty mathematics question is from the chapter integral calculus, covering the topic of definite integrals. It appeared in the 2025 exam.
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