Evaluate the integral:
∫ ( 2 − x ) 10 d x \int { { (2-x) }^{ 10 } } dx ∫ ( 2 − x ) 1 0 d x
Let us set u = 2 − x u=2-x u = 2 − x
Then d u d x = − 1 \dfrac { du }{ dx } =-1 d x d u = − 1 d u = − d x du=-dx d u = − d x d x = − d u dx=-du d x = − d u d u du d u d u d x \dfrac { du }{ dx } d x d u
Plugging u u u
∫ ( 2 − x ) 10 d x = ∫ u 10 ( − d u ) = − ∫ u 10 d u = − u 11 11 + C = − ( 2 − x ) 11 11 + C \int { { (2-x) }^{ 10 } } dx=\int { { u }^{ 10 }(-du) } =-\int { { u }^{ 10 }du } =-\frac { { u }^{ 11 } }{ 11 } +C=-\frac { { (2-x) }^{ 11 } }{ 11 } +C ∫ ( 2 − x ) 1 0 d x = ∫ u 1 0 ( − d u ) = − ∫ u 1 0 d u = − 1 1 u 1 1 + C = − 1 1 ( 2 − x ) 1 1 + C
Notice that the substitution made the integrand simpler, by making the problem into a problem where the power rule can be used.
Evaluate the integral:
∫ x 2 ( x 3 + 2 ) 101 d x \int { { x }^{ 2 }{ ({ x }^{ 3 }+2) }^{ 101 } } dx ∫ x 2 ( x 3 + 2 ) 1 0 1 d x
Let us set x 3 + 2 { x }^{ 3 }+2 x 3 + 2
Then d u = 3 x 2 d x du=3{ x }^{ 2 }dx d u = 3 x 2 d x d x = d u 3 x 2 dx=\dfrac { du }{ 3{ x }^{ 2 } } d x = 3 x 2 d u
Plugging u u u
∫ x 2 ( x 3 + 2 ) 101 d x = ∫ x 2 u 101 d u 3 x 2 = 1 3 ∫ u 101 d u = 1 3 u 102 102 + C = ( x 3 + 2 ) 102 306 + C \int { { x }^{ 2 }{ ({ x }^{ 3 }+2) }^{ 101 } } dx=\int { { x }^{ 2 }{ u }^{ 101 } } \frac { du }{ 3{ x }^{ 2 } } =\frac { 1 }{ 3 } \int { { u }^{ 101 } } du=\frac { 1 }{ 3 } \frac { { u }^{ 102 } }{ 102 } +C=\frac { { ({ x }^{ 3 }+2) }^{ 102 } }{ 306 } +C ∫ x 2 ( x 3 + 2 ) 1 0 1 d x = ∫ x 2 u 1 0 1 3 x 2 d u = 3 1 ∫ u 1 0 1 d u = 3 1 1 0 2 u 1 0 2 + C = 3 0 6 ( x 3 + 2 ) 1 0 2 + C
Notice that the substitution, again, made the integrand simpler, by making the problem into a problem where the power rule can be used.
Evaluate the integral:
∫ 1 1 − 3 x 4 d x \int { \frac { 1 }{ \sqrt [ 4 ]{ 1-3x } } } dx ∫ 4 1 − 3 x 1 d x
Let us set u = 1 − 3 x u=1-3x u = 1 − 3 x
Then d u = − 3 d x du=-3dx d u = − 3 d x d x = − d u 3 dx=-\dfrac { du }{ 3 } d x = − 3 d u
Plugging u u u
∫ 1 1 − 3 x 4 d x = ∫ 1 u 4 ( − d u 3 ) = − 1 3 ∫ 1 u 4 d u = − 1 3 ∫ u − 1 4 d u = − 1 3 u 3 4 3 4 + C = − 4 ( 1 − 3 x ) 3 4 9 + C \int { \frac { 1 }{ \sqrt [ 4 ]{ 1-3x } } } dx=\int { \frac { 1 }{ \sqrt [ 4 ]{ u } } } \left( -\frac { du }{ 3 } \right) =-\frac { 1 }{ 3 } \int { \frac { 1 }{ \sqrt [ 4 ]{ u } } } du=-\frac { 1 }{ 3 } \int { { u }^{ -\frac { 1 }{ 4 } } } du=-\frac { 1 }{ 3 } \frac { { u }^{ \frac { 3 }{ 4 } } }{ \frac { 3 }{ 4 } } +C=-\frac { 4\sqrt [ 4 ]{ { (1-3x) }^{ 3 } } }{ 9 } +C ∫ 4 1 − 3 x 1 d x = ∫ 4 u 1 ( − 3 d u ) = − 3 1 ∫ 4 u 1 d u = − 3 1 ∫ u − 4 1 d u = − 3 1 4 3 u 4 3 + C = − 9 4 4 ( 1 − 3 x ) 3 + C
Notice that the substitution, again, made the integrand simpler, by making the problem into a problem where the power rule can be used.
Evaluate the integral:
∫ sin 4 ( x ) cos ( x ) d x \int { \sin ^{ 4 }{ (x) } \cos { (x) } } dx ∫ sin 4 ( x ) cos ( x ) d x
Let us set u = sin ( x ) u=\sin { (x) } u = sin ( x )
Then d u = cos ( x ) d x du=\cos { (x) } dx d u = cos ( x ) d x d x = sec ( x ) d u dx=\sec { (x) } du d x = sec ( x ) d u
Plugging u u u
∫ sin 4 ( x ) cos ( x ) d x = ∫ u 4 cos ( x ) sec ( x ) d u = ∫ u 4 d u = u 5 5 + C = sin 5 ( x ) 5 + C \int { \sin ^{ 4 }{ (x) } \cos { (x) } } dx=\int { { u }^{ 4 }\cos { (x) } \sec { (x) } } du=\int { { u }^{ 4 } } du=\frac { { u }^{ 5 } }{ 5 } +C=\frac { \sin ^{ 5 }{ (x) } }{ 5 } +C ∫ sin 4 ( x ) cos ( x ) d x = ∫ u 4 cos ( x ) sec ( x ) d u = ∫ u 4 d u = 5 u 5 + C = 5 sin 5 ( x ) + C
Notice that the substitution made the integrand simpler, by making the problem into a problem where the derivative of sin ( x ) \sin { (x) } sin ( x )
Here is an exercise for you: Evaluate the integral:
∫ cos ( π x ) cos ( sin ( π x ) ) d x \int { \cos { (\pi x) } \cos { (\sin { (\pi x) } ) } } dx ∫ cos ( π x ) cos ( sin ( π x ) ) d x
Good Luck!
