Question

What is the derivative of `int_2^(cosx) \ sin^2(2t) \ dt`?

Answer 1

`" The Reqd. Deriv.="-sinx*sin^2(2cosx).`

Explanation:

Let, `f(x)=int_2^cosx sin^2(2t)dt.`

Now, `sin^2theta=(1-cos2theta)/2,` we have,

`f(x)=int_2^cosx sin^2(2t)dt=int_2^cosx (1-cos4t)/2 dt,`

`=1/2[t-1/4sin4t]_2^cosx,`

`=1/2[{cosx-1/4sin(4cosx)}-{2-1/4sin8}],`

`rArr f(x)=1/2{cosx-1/4sin(4cosx)}-1+1/8sin8.`

Therefore, the Reqd. Deriv.`=d/dxint_2^cosx sin^2(2t)dt.`

`=d/dx{f(x)},`

`=d/dx[1/2{cosx-1/4sin(4cosx)}-1+1/8sin8],`

`=1/2[d/dx cosx-d/dx 1/4sin(4cosx)]-0,`

`=1/2[-sinx-1/4cos(4cosx)*d/dx(4cosx),...[because," the Chain Rule]"`

`=-sinx/2-1/8cos(4cosx)*(-4sinx),`

`=-sinx/2+1/2sinxcos(4cosx),`

`=-sinx/2{1-cos(4cosx)}=-sinx/2(1-cos4theta),` say,

where, `theta=cosx,"`

`=-sinx/2(2sin^2 2theta),`

`rArr" the Reqd. Deriv.="-sinx*sin^2(2cosx).`

Enjoy Maths.!

Answer 2

`d/dx \ int_2^(cosx) \ sin^2(2t) \ dt = -sinx \ sin^2(2cosx)`

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

`d/dx \ int_a^x \ f(t) \ dt = f(x)` for any constant `a`

(ie the derivative of an integral gives us the original function back).

We are asked to find:

`d/dx \ int_2^(cosx) \ sin^2(2t) \ dt` ..... [A}

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

`u=cosx => (du)/dx = -sinx`

The substituting into the integral [A], and applying the chain rule, we get:

`d/dx \ int_2^(cosx) \ sin^2(2t) \ dt = d/dx \ int_2^u \ sin^2(2t) \ dt`

`" " = (du)/dx*d/(du) \ int_2^u \ sin^2(2t) \ dt`

`" " = -sinx \ d/(du) \ int_2^u \ sin^2(2t) \ dt`

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

`d/dx \ int_2^(cosx) \ sin^2(2t) \ dt = -sinx \ sin^2(2u)`

And restoring the initial substitution we get:

`d/dx \ int_2^(cosx) \ sin^2(2t) \ dt = -sinx \ sin^2(2cosx)`

And note that the given answer is incorrect!