0-4-sinx-cos-7-x-dx-solves- Tinku Tara June 3, 2023 Integration 0 Comments FacebookTweetPin Question Number 11436 by @ANTARES_VY last updated on 26/Mar/17 ∫π40sinx×cos7xdx.solves… Commented by FilupS last updated on 26/Mar/17 letu=cos(x)∴du=−sin(x)dx∴∫0π/4sin(x)cos7(x)dx=−∫x=0x=π/4u7du−∫x=0x=π/4u7du=−[18u8]x=0x=π/4=−18[cos8(x)]x=0x=π/4=−18(cos8(π4)−cos8(0))=−18((12)8−1)=−18(2−12×8−1)=−18(2−4−1)=−123(124−1)=−(127−123)=−(127−2427)=−(1−2427)=24−127=15128 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-142504Next Next post: In-a-ABC-triangle-the-side-a-6-and-c-2-b-2-66-Calculate-the-projections-of-sides-b-and-c-on-a- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![let u=cos(x) ∴ du=−sin(x)dx ∴∫_0 ^( π/4) sin(x)cos^7 (x)dx=−∫_(x=0) ^( x=π/4) u^7 du −∫_(x=0) ^( x=π/4) u^7 du=−[(1/8)u^8 ]_(x=0) ^(x=π/4) =−(1/8)[cos^8 (x)]_(x=0) ^(x=π/4) =−(1/8)(cos^8 ((π/4))−cos^8 (0)) =−(1/8)(((1/( (√2))))^8 −1) =−(1/8)(2^(−(1/2)×8) −1) =−(1/8)(2^(−4) −1) =−(1/2^3 )((1/2^4 )−1) =−((1/2^7 )−(1/2^3 )) =−((1/2^7 )−(2^4 /2^7 )) =−(((1−2^4 )/2^7 )) =((2^4 −1)/2^7 ) =((15)/(128))](https://www.tinkutara.com/question/Q11442.png)