Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 209709 by Ismoiljon_008 last updated on 19/Jul/24

$$$$$$\int {(sinx+cosx)}^{11}dx=?$$$$helpmeplease$$

Answered by mr W last updated on 19/Jul/24

$$\int {(\sin x+\cos x)}^{11}dx$$$$=\int {\left(\sqrt{2}\cos (x-\frac{\pi }{4})\right)}^{11}dx$$$$=32\sqrt{2}\int {\cos }^{11}(x-\frac{\pi }{4})d(x-\frac{\pi }{4})$$$$=32\sqrt{2}\int {\cos }^{11}udu$$$$=32\sqrt{2}\int {\cos }^{10}ud\left(\sin u\right)$$$$=32\sqrt{2}\int {(1-{\sin }^{2}u)}^{5}d\left(\sin u\right)$$$$=32\sqrt{2}\int {(1-t^2)}^{5}dt$$$$=32\sqrt{2}\int (1-5t^2+10t^4-10t^6+5t^8-t^{10})dt$$$$=32\sqrt{2}(t-\frac{5t^3}{3}+2t^5-\frac{10t^7}{7}+\frac{5t^9}{9}-\frac{t^{11}}{11})+C$$$$=32\sqrt{2}[\sin (x-\frac{\pi }{4})-\frac{5{\sin }^3(x-\frac{\pi }{4})}{3}+2{\sin }^5(x-\frac{\pi }{4})-\frac{10{\sin }^7(x-\frac{\pi }{4})}{7}+\frac{5{\sin }^9(x-\frac{\pi }{4})}{9}-\frac{{\sin }^{11}(x-\frac{\pi }{4})}{11}]+C$$

Commented by Ismoiljon_008 last updated on 19/Jul/24

$$thankyouverymuch$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com