# How I calculate these integrals without calculator?

### 1 Answer | Add Yours

(1) `int u3^udu`

To solve, apply the integration by parts.

`int UdV= UV - intVdU`

In the given integral, let our U and dV be:

`U=u` and `dV=3^udu`

Then, differentiate U to get dU.

`dU=1du`

`dU=du`

Then, take the integral of dV to get V.

`int dV = int 3^udu`

`V=3^u/(ln3)`

So we have `U=u` , `dU=du` , `V=3^u/(ln3)` and `dV=3^udu` .

Now that we know our U, dU and V, plug-in them to the formula of integration by parts.

`int UdV = UV - intVdU`

`int u3^udu = u*3^u/(ln3) - int 3^u/(ln3)du`

`int u3^udu = (u3^u)/(ln3) - 1/(ln3)int3^udu`

`int u3^udu = (u3^u)/(ln3)-1/(ln3)*3^u/(ln3)+C`

`int u3^udu = (u3^u)/(ln3)-3^u/(ln3)^2 + C`

**Therefore, `int u3^udu = (u3^u)/(ln3)-3^u/(ln3)^2 + C ` .**

(2) `int_0^(pi/2) |sinx-cos(2x)|dx`

To integrate this, we have to remove the absolute value sign. To do this, apply the definition of absolute value which is:

`|x| = {(x if xgt=0), (-x if xlt0):}`

That means when we remove the absolute value sign, we have to split our integral into two, depending on where the sinx - cos(2x) is positive.

Since the limit of the integral is from 0 to pi/2, we have to determine the values of x in that interval in which the expression inside the absolute value is always positive. To do so, set sinx -cos(2x) greater than and equal to zero.

`sinx - cos(2x) gt=0`

`sinx-(1-2sin^2x)gt=0`

`2sin^2x + sinx -1gt=0`

`(2sinx-1)(sinx + 1)gt=0`

Applying the method of solving inequality equation, for the interval [0, pi/2], sinx - cos(2x) is always positive when the value of x is from pi/6 to pi/2 `(pi/6lt=x<=pi/2)` .

Now that we know the values of x in which sin(x) - cos(2x) is positive, then, using the definition of absolute value, we may express |sinx - cos(2x)| as:

`|sinx-cos(2x)|={(sinx - cos(2x) if pi/6lt=xlt=pi/2), (-(sinx-cos(2x)) if 0lt=xltpi/6):}`

So removing the absolute value sign, the integral becomes:

`int_0^(pi/2) |sinx-cos(2x)|dx`

`= int_0^(pi/6) -(sinx-cos(2x))dx + int_(pi/6)^(pi/2) (sinx - cos(2x))dx`

`=-int_0^(pi/6)(sinx-cos(2x))dx + int_(pi/6)^(pi/2) (sinx - cos(2x))dx`

`=-(-cosx - sin(2x)/2)|_0^(pi/6) + (-cosx - sin(2x)/2)_(pi/6)^(pi/2)`

`=(cosx + sin(2x)/2)|_0^(pi/6) + (-cosx - sin(2x)/2)|_(pi/6)^(pi/2)`

`=[(sqrt3/2 +sqrt3/4) - (1+0)] + [(0-0) - (-sqrt3/2-sqrt3/4)]`

`=[ (3sqrt3)/4 - 1] + [ 0 -(-3sqrt3/4)]`

`=(3sqrt3)/4-1+(3sqrt3)/4`

`=(6sqrt3)/4-1`

`=(3sqrt3)/2-1`

**Thus, `int_0^(pi/2) |sinx-cos(2x)|dx=(3sqrt3)/2-1` .**