simplify this: \int \frac{\cos^{-1}(x) \cdot \sqrt{(1-x^2)^{-1}}}{\log \left( 1+ \left( \frac{\sin (2x\sqrt{1-x^2})}{\pi} \right) \right)} \, dx

Question

Bytelearn · Accepted Answer

Analyze the integral: Analyze the integral expression. We have the integral expression: $$\int \frac{\cos^{-1}(x) \cdot \sqrt{(1-x^2)^{-1}}}{\log \left( 1+ \left( \frac{\sin (2x\sqrt{1-x^2})}{\pi} \right) \right)} \, dx$$ We need to simplify this expression. First, let's look at the square root in the numerator.Simplify square root: Simplify the square root in the numerator. The square root of the inverse is the same as the square root of the reciprocal, so we have: $$\sqrt{(1-x^2)^{-1}} = \frac{1}{\sqrt{1-x^2}}$$ Now, the integral expression becomes: $$\int \frac{\cos^{-1}(x)}{\sqrt{1-x^2} \cdot \log \left( 1+ \left( \frac{\sin (2x\sqrt{1-x^2})}{\pi} \right) \right)} \, dx$$Recognize derivative: Recognize the derivative of the inverse cosine function. The derivative of $\cos^{-1}(x)$ is $-\frac{1}{\sqrt{1-x^2}}$. This is similar to the expression we have in the numerator. However, we have a positive sign instead of a negative sign, which we need to account for.Adjust integral: Adjust the integral to match the derivative of the inverse cosine function. To make the numerator match the derivative of $\cos^{-1}(x)$, we can multiply and divide by -1: $$\int \frac{-1 \cdot \cos^{-1}(x)}{-\sqrt{1-x^2} \cdot \log \left( 1+ \left( \frac{\sin (2x\sqrt{1-x^2})}{\pi} \right) \right)} \, dx$$ Now, the numerator is the derivative of $\cos^{-1}(x)$ times -1.Look for substitution: Look for a substitution. We can let $u = \cos^{-1}(x)$, which implies $x = \cos(u)$. Then, $dx = -\sin(u) \, du$, and we have $\sqrt{1-x^2} = \sin(u)$.Perform the substitution: Perform the substitution. Substituting $u$ and $dx$ into the integral, we get: $$\int \frac{-u}{-\sin(u) \cdot \log \left( 1+ \left( \frac{\sin (2\cos(u)\sin(u))}{\pi} \right) \right)} \cdot (-\sin(u) \, du)$$ This simplifies to: $$\int \frac{u}{\log \left( 1+ \left( \frac{\sin (2\cos(u)\sin(u))}{\pi} \right) \right)} \, du$$Simplify sine argument: Simplify the argument of the sine function in the denominator. The argument of the sine function inside the logarithm is $2\cos(u)\sin(u)$, which is equivalent to $\sin(2u)$ due to the double-angle formula for sine.Substitute double-angle formula: Substitute the double-angle formula into the integral. The integral now becomes: $$\int \frac{u}{\log \left( 1+ \left( \frac{\sin(2u)}{\pi} \right) \right)} \, du$$Recognize complexity: Recognize the complexity of the integral. At this point, we have an integral that involves a logarithm in the denominator with a non-trivial argument. This integral does not have a standard form and cannot be easily simplified using elementary functions. It may require special functions or numerical methods to solve.Conclude integral: Conclude that the integral cannot be simplified further in terms of elementary functions. The integral: $$\int \frac{u}{\log \left( 1+ \left( \frac{\sin(2u)}{\pi} \right) \right)} \, du$$ does not simplify to a form expressible in terms of elementary functions. Therefore, we cannot provide a simplified expression for the original integral.

simplify this: $\int \frac{\cos^{-1}(x) \cdot \sqrt{(1-x^2)^{-1}}}{\log \left( 1+ \left( \frac{\sin (2x\sqrt{1-x^2})}{\pi} \right) \right)} \, dx$

Full solution

More problems from Multiplication with rational exponents

Related topics

simplify this: ∫cos⁡−1(x)⋅(1−x2)−1log⁡(1+(sin⁡(2x1−x2)π)) dx\int \frac{\cos^{-1}(x) \cdot \sqrt{(1-x^2)^{-1}}}{\log \left( 1+ \left( \frac{\sin (2x\sqrt{1-x^2})}{\pi} \right) \right)} \, dx∫log(1+(πsin(2x1−x2​)​))cos−1(x)⋅(1−x2)−1​​dx

simplify this: $\int \frac{\cos^{-1}(x) \cdot \sqrt{(1-x^2)^{-1}}}{\log \left( 1+ \left( \frac{\sin (2x\sqrt{1-x^2})}{\pi} \right) \right)} \, dx$