Does my work look accurate. Wasnt sure if i could take the 4 out at the last step.
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Sep 21, 2010 #1 Does my work look accurate. Wasnt sure if i could take the 4 out at the last step.
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Sep 21, 2010 #2 Evaluate ∫0π/85cos(4t)sin(4t)dt\displaystyle Evaluate \ \int_{0}^{\pi/8}5^{cos(4t)}sin(4t)dtEvaluate ∫0π/85cos(4t)sin(4t)dt Let u = cos(4t), then du = −4sin(4t)dt, ⟹ sin(4t)dt = du−4\displaystyle Let \ u \ = \ cos(4t), \ then \ du \ = \ -4sin(4t)dt, \ \implies \ sin(4t)dt \ = \ \frac{du}{-4}Let u = cos(4t), then du = −4sin(4t)dt, ⟹ sin(4t)dt = −4du Hence, −14∫105udu = 14∫015udu = 5u4ln(5)]01\displaystyle Hence, \ \frac{-1}{4}\int_{1}^{0}5^udu \ = \ \frac{1}{4}\int_{0}^{1}5^udu \ = \ \frac{5^u}{4ln(5)}\bigg]_{0}^{1}Hence, 4−1∫105udu = 41∫015udu = 4ln(5)5u]01 = 54ln(5)−14ln(5) = 1ln(5)\displaystyle = \ \frac{5}{4ln(5)}-\frac{1}{4ln(5)} \ = \ \frac{1}{ln(5)}= 4ln(5)5−4ln(5)1 = ln(5)1
Evaluate ∫0π/85cos(4t)sin(4t)dt\displaystyle Evaluate \ \int_{0}^{\pi/8}5^{cos(4t)}sin(4t)dtEvaluate ∫0π/85cos(4t)sin(4t)dt Let u = cos(4t), then du = −4sin(4t)dt, ⟹ sin(4t)dt = du−4\displaystyle Let \ u \ = \ cos(4t), \ then \ du \ = \ -4sin(4t)dt, \ \implies \ sin(4t)dt \ = \ \frac{du}{-4}Let u = cos(4t), then du = −4sin(4t)dt, ⟹ sin(4t)dt = −4du Hence, −14∫105udu = 14∫015udu = 5u4ln(5)]01\displaystyle Hence, \ \frac{-1}{4}\int_{1}^{0}5^udu \ = \ \frac{1}{4}\int_{0}^{1}5^udu \ = \ \frac{5^u}{4ln(5)}\bigg]_{0}^{1}Hence, 4−1∫105udu = 41∫015udu = 4ln(5)5u]01 = 54ln(5)−14ln(5) = 1ln(5)\displaystyle = \ \frac{5}{4ln(5)}-\frac{1}{4ln(5)} \ = \ \frac{1}{ln(5)}= 4ln(5)5−4ln(5)1 = ln(5)1
R Ryan Rigdon Junior Member Joined Jun 10, 2010 Messages 246 Sep 21, 2010 #3 like how you did yours BigGlenntheHeavy. seems easier than what i did. thanx again for your work.
