`y=int_7^(tan(x))sqrt(3t+sqrt(t)))dt`

The first part of the theorem says,if,

`F(x) = int_a^xf(t)dt` then,

`F'(x) = f(x)`

In our case, we have tan x instead of x, so we can't apply this theorem straight away, so we need a substitution. we will use

`u =tan(x) `

so , `(du)/(dx) = sec^2(x)`

now...

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`y=int_7^(tan(x))sqrt(3t+sqrt(t)))dt`

The first part of the theorem says,if,

`F(x) = int_a^xf(t)dt` then,

`F'(x) = f(x)`

In our case, we have tan x instead of x, so we can't apply this theorem straight away, so we need a substitution. we will use

`u =tan(x) `

so , `(du)/(dx) = sec^2(x)`

now our integral becomes,

`y=int_7^usqrt(3t+sqrt(t)))dt`

but we need '0' on the lower limit.

`y=int_0^usqrt(3t+sqrt(t)))dt - int_0^7sqrt(3t+sqrt(t))dt`

so,

dy/du becomes,

`(dy)/(du) = f(u) - f(7)`

`(dy)/(du) = sqrt(3u+sqrt(u)) - sqrt(3*7+sqrt(7))`

`(dy)/(du) = sqrt(3u+sqrt(u)) - 4.862689`

But we know as per chain rule,

`(dy)/(dx) = (dy)/(du) * (du)/(dx)`

so we get,

`(dy)/(dx) = (sqrt(3u+sqrt(u)) - 4.862689) * sec^2(x)`

but u is tan(x)

`(dy)/(dx) = (sqrt(3tan(x)+sqrt(tan(x))) - 4.862689) * sec^2(x)`