# How do you integrate the (x*sqrt(16+x^2) )dx from 0 to 3? `int_0^3xsqrt(16+x^2)dx` please help, I'm stuck on it

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### 1 Answer

`y = int_0^3xsqrt(16+x^2)dx`

This can be done by using a trigonometric substiution.

Let `x = 4tan(t)`

Then, `dx = 4sec^2(t)dt`

When `x = 0` ,

`0 = 4tan(t)`

`tan(t) = 0`

`t = 0`

When `x =3, `

`3 = 4tan(t)`

`t = tan^(-1)(3/4) = alpha`

`sec(alpha) = sqrt(1 + tan^2(alpha)) = sqrt(1 + 9/16) = sqrt(25/16)`

`sec(alpha) = 5/4`

Now we can change the integral as,

`y = int_0^alpha 4tan(t)sqrt(16+16tan^2(t)) xx 4sec^2(t)dt`

`y = int_0^alpha 4tan(t) xx4 xxsqrt(1+tan^2(t)) xx 4sec^2(t)dt`

`y = int_0^alpha 64 tan(t) xx sec(t) xx sec^2(t) dt`

This gives,

`y = 64int_0^alphasec^2(t)sec(t)tan(t)dt`

We know, that,

`d(sec(t)) = sec(t)tan(t)dt`

Therefore, integral operator changes into,

`y = 64int_0^alphasec^2(t)d(sec(t))`

Now integrating wrt `sec(t)` ,

`y = 64 [1/3sec^3(t)]_0^alpha`

`y = 64/3 [sec^(alpha) -sec^3(0)]`

`y = 64/3[(5/4)^3-1]`

`y = 64/3(125/64 -1)`

`y = 61/3`

Therefore,

`y = int_0^3xsqrt(16+x^2)dx = 61/3`

**Through calculator I got 20.333. Therefore this is correct.**