A.)∫10x3x4+1dx
Let us assume that:
x4+1=u⇒4x3dx=du
⇒x3dx=4du
When:
x=1
Then:
u=14+1=2
Again, when:
x=0
Then:
u=04+1=1
Substituting u=x4+1 into the integration and integrating with respect to du , we have:
=41∫21udu
=41∫21u21du
=41[23u23]21 [∵∫xndx=n+1xn+1]
=61[u23]21
=61[123−223]=61−22
B.) ∫4π0tan7(x)sec2(x)dx
Let us assume that:
tan(x)=u⇒sec2(x)dx=du
When:
x=4π
Then:
u=tan(4π)=1
Again, when:
x=0
Then:
u=tan(0)=0
Substituting u=tan(x) into the integration and integrating with respect to du , we have:
=∫10u7du=81[u8]10 [∵∫xndx=n+1xn+1]
=−81
C.) ∫sec2(cos(x))sin(x)dx
Let us assume that:
cos(x)=u⇒−sin(x)dx=du
Substituting u=cos(x) into the integration and integrating with respect to du , we have:
=−∫sec2(u)du=−tan(u)+C [∵∫sec2(x)dx=tan(x)+C]
Where C is a constant of integration.
Undo substitution and we have:
=−tan(cos(x))+C [∵u=cos(x)]
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