integrate from 0 to pi/4 (sec^2x)/((1+7tanx)^2)^1/3
integrate form pi^2/36 to pi^2/4 (cos(x^1/2))/(xsin(x^1/2))^1/2
integrate from 0 to pi/3 (tanx)/(2secx)^1/2
if we let
u = 1+7tanx
du = 7sec^2 x dx
and you have ∫ u^(-1/3) 1/7 du
if we let
u = sin√x
du = cos√x * 1/(2√x)
and you have ∫ 1/√u 2 du
To solve these integrals, we will need to use some integration techniques and properties. Let's break down each integral and see how we can proceed.
1. ∫[0, π/4] [(sec^2x)/((1+7tanx)^2)^(1/3)] dx:
To start, notice that sec^2(x) = 1 + tan^2(x). We can rewrite the integrand as:
∫[0, π/4] [(1 + tan^2(x))/((1+7tanx)^2)^(1/3)] dx.
Now, let's substitute u = 1 + 7tan(x):
du/dx = 7sec^2(x) => dx = du/(7sec^2(x)) = du/(7(u-1)).
Using this substitution, the integral becomes:
∫[(u=1) to (u=2)] (1 + tan^2(x))/(u^2)^(1/3) * du/(7(u-1)).
Simplifying further:
∫[(u=1) to (u=2)] ((1 + (u-1)/7)^2)/(u^2)^(1/3) * du/(7(u-1)).
Now, let's simplify the expression inside the integral:
(1 + (u-1)/7)^2 = (u/7)^2 = (u^2)/49.
Next, simplify (u^2)^(1/3):
(u^2)^(1/3) = u^(2/3).
Substituting everything back in, the integral simplifies to:
(1/49) * ∫[(u=1) to (u=2)] (u^(2/3))/(u^2-2u) du.
Now, we can split the numerator into two separate terms:
(1/49) * ∫[(u=1) to (u=2)] (u^(-1/3))/(u-2) du + (1/49) * ∫[(u=1) to (u=2)] (u^(5/3))/(u-2) du.
These integrals are now in a form that can be solved using the power rule of integration. Finally, evaluate the integrals over the given range to find the solution.
2. ∫[π^2/36, π^2/4] (cos(x^(1/2)))/(xsin(x^(1/2)))^(1/2) dx:
Let's start by substituting u = x^(1/2):
du/dx = (1/2)x^(-1/2) => dx = 2udu.
Using this substitution, the integral becomes:
∫[(u=π/6) to (u=π/2)] (cos(u))/(2u*sin(u))^(1/2) * 2udu.
Simplifying further:
∫[(u=π/6) to (u=π/2)] (cos(u))/(u*sin(u))^(1/2) du.
This integral can be solved using a method called integration by parts. Let's denote the integrand as f(u) and choose the following u and dv:
u = (u*sin(u))^(1/2) => du = (2sin(u) + u*cos(u))/(2(u*sin(u))^(1/2)) du,
dv = (cos(u))/u.
Now, let's find du and v:
du = (2sin(u) + u*cos(u))/(2(u*sin(u))^(1/2)) du,
v = ∫(cos(u))/u du = Ci(u) (cosine integral).
Using the integration by parts formula, the integral becomes:
∫[(u=π/6) to (u=π/2)] (cos(u))/(u*sin(u))^(1/2) du = [(u*sin(u))^(1/2) * Ci(u)] - ∫[(π/6) to (π/2)] Ci(u) * (2sin(u) + u*cos(u))/(2(u*sin(u))^(1/2)) du.
This integral might not have a closed-form solution, but you can use numerical approximation methods to find an approximate value.
3. ∫[0, π/3] (tan(x))/(2sec(x))^(1/2) dx:
First, let's simplify (2sec(x))^(1/2):
(2sec(x))^(1/2) = (2/cos(x))^(1/2) = (2^(1/2))/(cos(x))^(1/2).
Now, the integral becomes:
∫[0, π/3] tan(x) * (2^(1/2) * (cos(x))^(1/2)) dx.
Using the identity sec^2(x) = 1 + tan^2(x), we can rewrite the integral as:
∫[0, π/3] (tan^2(x) + tan(x))/(cos(x))^(1/2) dx.
Let's make a substitution: u = cos(x).
du/dx = -sin(x) => dx = -du/sin(x) = -du/(√(1-u^2)).
Substituting everything back in, the integral becomes:
∫[(u=1) to (u=√3/2)] (tan^2(x) + tan(x))/(u)^(1/2) * -du/(√(1-u^2)).
This integral can be solved by using the power rule for integration. Evaluate the integral over the given range to find the solution.