Integral of sin inverse x into log x

To find the integral of sin inverse x into log x, we can use integration by parts. Integration by parts is a technique used to evaluate integrals by rewriting them in a different form. The formula for integration by parts is given by:

∫u dv = uv - ∫v du

Where u and v are functions of x, and du and dv are their respective differentials.

Step 1: Choose u and dv
In this case, we can choose:
u = log x
dv = sin inverse x dx

Step 2: Find du and v
To find du and v, we need to differentiate u and integrate dv respectively.

Differentiating u:
du = 1/x dx

Integrating dv:
v = ∫sin inverse x dx

Step 3: Evaluate du and v
To evaluate du and v, we need to substitute the values of du and v obtained from the previous step.

Substituting du and v:
du = 1/x dx
v = ∫sin inverse x dx

Step 4: Apply the integration by parts formula
Using the integration by parts formula, we can rewrite the integral as:

∫u dv = uv - ∫v du

Substituting the values, we get:

∫log x (sin inverse x) dx = (log x)(∫sin inverse x dx) - ∫(∫sin inverse x dx)(1/x) dx

Simplifying further:

∫log x (sin inverse x) dx = (log x)(∫sin inverse x dx) - ∫(1/x)(∫sin inverse x dx) dx

Step 5: Evaluate the integrals
Now, we need to evaluate the remaining integrals. The integral of sin inverse x can be found using integration by substitution or by using trigonometric identities. The integral of (1/x) can be evaluated using logarithmic properties.

Once the integrals are evaluated, we can substitute the values back into the equation to find the integral of sin inverse x into log x.

Conclusion
To find the integral of sin inverse x into log x, we used integration by parts. By choosing appropriate functions for u and dv and applying the integration by parts formula, we obtained an expression involving the integral of sin inverse x and the integral of (1/x). Evaluating these integrals will give us the final result.