integral of sin^3(17x)cos^3(17x)dx[/code]
P peepsqueeq New member Joined May 24, 2007 Messages 1 May 24, 2007 #1 integral of sin^3(17x)cos^3(17x)dx[/code]
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 May 24, 2007 #2 Hello, peepsqueeq1 \(\displaystyle \L\int\)sin3(17x)⋅cos3(17x) dx\displaystyle \sin^3(17x)\cdot\cos^3(17x)\,dxsin3(17x)⋅cos3(17x)dx Click to expand... We have: \(\displaystyle \L\:\int\)sin3(17x)⋅cos2(17x)⋅cos(17x) dx\displaystyle \sin^3(17x)\cdot\cos^2(17x)\cdot\cos(17x)\,dxsin3(17x)⋅cos2(17x)⋅cos(17x)dx . . . \(\displaystyle =\;\L\int\)sin3(17x)⋅[1 − sin2(17x)]⋅cos(17x) dx\displaystyle \sin^3(17x)\cdot\left[1\,-\,\sin^2(17x)\right]\cdot\cos(17x)\,dxsin3(17x)⋅[1−sin2(17x)]⋅cos(17x)dx . . . \(\displaystyle = \;\L\int\)[sin3(17x) − sin5(17x)]⋅cos(17x) dx\displaystyle \left[\sin^3(17x)\,-\,\sin^5(17x)\right]\cdot\cos(17x)\,dx[sin3(17x)−sin5(17x)]⋅cos(17x)dx Let u = sin(17x) ⇒ du = 17⋅cos(17x) dx ⇒ cos(17x) dx = 117 du\displaystyle u\:=\:\sin(17x)\;\;\Rightarrow\;\;du\:=\:17\cdot\cos(17x)\,dx\;\;\Rightarrow\;\;\cos(17x)\,dx \:=\:\frac{1}{17}\,duu=sin(17x)⇒du=17⋅cos(17x)dx⇒cos(17x)dx=171du Substitute: \(\displaystyle \L\:\int\)\(\displaystyle \left(u^3\,-\,u^5\right)\left(\frac{1}{17}\,du\right)\;=\;\frac{1}{17}\L\int\)(u3 − u5) du\displaystyle \left(u^3\,-\,u^5\right)\,du(u3−u5)du Can you finish it now?
Hello, peepsqueeq1 \(\displaystyle \L\int\)sin3(17x)⋅cos3(17x) dx\displaystyle \sin^3(17x)\cdot\cos^3(17x)\,dxsin3(17x)⋅cos3(17x)dx Click to expand... We have: \(\displaystyle \L\:\int\)sin3(17x)⋅cos2(17x)⋅cos(17x) dx\displaystyle \sin^3(17x)\cdot\cos^2(17x)\cdot\cos(17x)\,dxsin3(17x)⋅cos2(17x)⋅cos(17x)dx . . . \(\displaystyle =\;\L\int\)sin3(17x)⋅[1 − sin2(17x)]⋅cos(17x) dx\displaystyle \sin^3(17x)\cdot\left[1\,-\,\sin^2(17x)\right]\cdot\cos(17x)\,dxsin3(17x)⋅[1−sin2(17x)]⋅cos(17x)dx . . . \(\displaystyle = \;\L\int\)[sin3(17x) − sin5(17x)]⋅cos(17x) dx\displaystyle \left[\sin^3(17x)\,-\,\sin^5(17x)\right]\cdot\cos(17x)\,dx[sin3(17x)−sin5(17x)]⋅cos(17x)dx Let u = sin(17x) ⇒ du = 17⋅cos(17x) dx ⇒ cos(17x) dx = 117 du\displaystyle u\:=\:\sin(17x)\;\;\Rightarrow\;\;du\:=\:17\cdot\cos(17x)\,dx\;\;\Rightarrow\;\;\cos(17x)\,dx \:=\:\frac{1}{17}\,duu=sin(17x)⇒du=17⋅cos(17x)dx⇒cos(17x)dx=171du Substitute: \(\displaystyle \L\:\int\)\(\displaystyle \left(u^3\,-\,u^5\right)\left(\frac{1}{17}\,du\right)\;=\;\frac{1}{17}\L\int\)(u3 − u5) du\displaystyle \left(u^3\,-\,u^5\right)\,du(u3−u5)du Can you finish it now?
