integration by substitution (1 Viewer)

Masaken · Jan 18, 2023

how do i apply the substitution here? no idea what to use for the u substitution

i tried to simplify the expression with trig identities but i'm not sure where to go next

thanks in advance

5uckerberg · Jan 18, 2023

Masaken said:
View attachment 37467
how do i apply the substitution here? no idea what to use for the u substitution

i tried to simplify the expression with trig identities but i'm not sure where to go next

thanks in advance

Is this the first time you saw this question? If so then I understand why this question felt a bit difficult.

Here let us start.

Step 1: $\int{\sin^{2}x\cos^{3}xdx}=\int{\sin^{2}x\cos^{2}x\cdot{\cos{x}}dx}$
Step 2: $\int{\sin^{2}x\cos^{2}x\cdot{\cos{x}}dx}=\int{\sin^{2}x\left(1-\sin^{2}x\right)\cdot{\cos{x}}dx}$
Step 3: $\int{\sin^{2}x\left(1-\sin^{2}x\right)\cdot{\cos{x}}dx}=\int{\sin^{2}x\cos{x}dx}-\int{\sin^{4}x\cos{x}dx}$
Step 4: $\int{\sin^{2}x\cos{x}dx}-\int{\sin^{4}x\cos{x}dx}$ Let $u=\sin{x}, du=\cos{x}dx$
Step 5: $... = \frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C$

gamja · Jan 18, 2023

Masaken said:
View attachment 37467
how do i apply the substitution here? no idea what to use for the u substitution

i tried to simplify the expression with trig identities but i'm not sure where to go next

thanks in advance

like 5uckerberg's correct answer, the general theory behind trig u-sub integration is:

If there are all even powers of trig functions, e.g. $\int \sin ^2\left(x\right)\cos ^2\left(x\right)dx$ , then convert everything into cos double angle form, factorise and work from there
If there are one or more ODD powers of trig functions, e.g. $\int \sin ^2\left(x\right)\cos ^3\left(x\right)dx$ , then we need to grab a sin or a cos as the f'(x) and use sin^x+cos^x=1 identity to make everything else be powers of one trig function, so our reverse chain rule works out.

Things get a little funkier with tan/sec integration but it's essentially the same as principle #2.

hope that kinda helps lol

Drongoski · Jan 18, 2023

Without the formal process of substitution. Of course, if question asks you to use the method of substitution, then you must use the method.

$\int sin^2 x cos^3 x dx = \int sin^2 x \times cos^2 x \times cos x dx = \int sin^2x (1-sin^2x)dsinx \\ \\ = \int ((sin x)^2 - (sin x)^4) dsin x = \frac {1}{3} sin^3 x - \frac {1}{5} sin^5 x + C$

Masaken · Jan 18, 2023

5uckerberg said:
Is this the first time you saw this question? If so then I understand why this question felt a bit difficult.

Here let us start.

Step 1: $\int{\sin^{2}x\cos^{3}xdx}=\int{\sin^{2}x\cos^{2}x\cdot{\cos{x}}dx}$
Step 2: $\int{\sin^{2}x\cos^{2}x\cdot{\cos{x}}dx}=\int{\sin^{2}x\left(1-\sin^{2}x\right)\cdot{\cos{x}}dx}$
Step 3: $\int{\sin^{2}x\left(1-\sin^{2}x\right)\cdot{\cos{x}}dx}=\int{\sin^{2}x\cos{x}dx}-\int{\sin^{4}x\cos{x}dx}$
Step 4: $\int{\sin^{2}x\cos{x}dx}-\int{\sin^{4}x\cos{x}dx}$ Let $u=\sin{x}, du=\cos{x}dx$
Step 5: $... = \frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C$

yea this was my first time, whoops. makes so much more sense to put in the substitution later, i was trying to do it at the beginning
thanks a lot!!

gamja said:
like 5uckerberg's correct answer, the general theory behind trig u-sub integration is:

If there are all even powers of trig functions, e.g. $\int \sin ^2\left(x\right)\cos ^2\left(x\right)dx$ , then convert everything into cos double angle form, factorise and work from there

If there are one or more ODD powers of trig functions, e.g. $\int \sin ^2\left(x\right)\cos ^3\left(x\right)dx$ , then we need to grab a sin or a cos as the f'(x) and use sin^x+cos^x=1 identity to make everything else be powers of one trig function, so our reverse chain rule works out.

Things get a little funkier with tan/sec integration but it's essentially the same as principle #2.

hope that kinda helps lol

this was very helpful, thanks!!

Masaken · Jan 18, 2023

Drongoski said:
Without the formal process of substitution. Of course, if question asks you to use the method of substitution, then you must use the method.

$\int sin^2 x cos^3 x dx = \int sin^2 x cos^2 x cos x dx = \int sin^2x (1-sin^2x)dsinx\\ \\ = \int ((sin x)^2 - (sin x)^4) dsin x = \frac {1}{3} sin^3 x - \frac {1}{5} sin^5 x + C$

thanks!! that makes sense

integration by substitution (1 Viewer)

Masaken

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5uckerberg

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gamja

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Drongoski

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Masaken

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Masaken

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