Let I1=∫aπ−a xf(sin⁡x)dx,I2=∫aπ−a f(sin⁡x)dx, then I2 is equal to 

 Let I1=aπaxf(sinx)dx,I2=aπaf(sinx)dx, then I2 is equal to 

  1. A

    π2I1

  2. B

    πI1

  3. C

    2πI1

  4. D

    2I1

    Register to Get Free Mock Test and Study Material

    +91

    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

     GivenI1=aπaxf(sinx)dxand I2=aπaf(sinx)dx

        Now,  I1=aπaxf(sinx)dx=aπa(πx)f[sin(πx)]dx=aπa(πx)f(sinx)dx=aπaπf(sinx)dxI1 2I1=πI2 I2=2πI1

    Chat on WhatsApp Call Infinity Learn

      Register to Get Free Mock Test and Study Material

      +91

      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.