Solve ∫ wrt x/1+frac{1(frac{pi{2}-x)}{Q(pi/2-x)}}

$$\int \frac { d x } { 1 + \frac { 1 ( \frac { \pi } { 2 } - x ) } { Q ( \frac { \pi } { 2 } - x ) } }$$

Evaluate

$$\frac{Qx}{Q+1}+С$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2}{2}.

$$\int \frac{1}{1+\frac{1\left(\frac{\pi }{2}-\frac{2x}{2}\right)}{Q\left(\frac{\pi }{2}-x\right)}}\mathrm{d}x$$

Since \frac{\pi }{2} and \frac{2x}{2} have the same denominator, subtract them by subtracting their numerators.

$$\int \frac{1}{1+\frac{1\times \frac{\pi -2x}{2}}{Q\left(\frac{\pi }{2}-x\right)}}\mathrm{d}x$$

Express 1\times \frac{\pi -2x}{2} as a single fraction.

$$\int \frac{1}{1+\frac{\frac{\pi -2x}{2}}{Q\left(\frac{\pi }{2}-x\right)}}\mathrm{d}x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2}{2}.

$$\int \frac{1}{1+\frac{\frac{\pi -2x}{2}}{Q\left(\frac{\pi }{2}-\frac{2x}{2}\right)}}\mathrm{d}x$$

Since \frac{\pi }{2} and \frac{2x}{2} have the same denominator, subtract them by subtracting their numerators.

$$\int \frac{1}{1+\frac{\frac{\pi -2x}{2}}{Q\times \frac{\pi -2x}{2}}}\mathrm{d}x$$

Express Q\times \frac{\pi -2x}{2} as a single fraction.

$$\int \frac{1}{1+\frac{\frac{\pi -2x}{2}}{\frac{Q\left(\pi -2x\right)}{2}}}\mathrm{d}x$$

Divide \frac{\pi -2x}{2} by \frac{Q\left(\pi -2x\right)}{2} by multiplying \frac{\pi -2x}{2} by the reciprocal of \frac{Q\left(\pi -2x\right)}{2}.

$$\int \frac{1}{1+\frac{\left(\pi -2x\right)\times 2}{2Q\left(\pi -2x\right)}}\mathrm{d}x$$

Cancel out 2\left(-2x+\pi \right) in both numerator and denominator.

$$\int \frac{1}{1+\frac{1}{Q}}\mathrm{d}x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{Q}{Q}.

$$\int \frac{1}{\frac{Q}{Q}+\frac{1}{Q}}\mathrm{d}x$$

Since \frac{Q}{Q} and \frac{1}{Q} have the same denominator, add them by adding their numerators.

$$\int \frac{1}{\frac{Q+1}{Q}}\mathrm{d}x$$

Divide 1 by \frac{Q+1}{Q} by multiplying 1 by the reciprocal of \frac{Q+1}{Q}.

$$\int \frac{Q}{Q+1}\mathrm{d}x$$

Find the integral of \frac{Q}{Q+1} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{Q}{Q+1}x$$

Simplify.

$$\frac{Qx}{Q+1}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{Qx}{Q+1}+С$$

Differentiate w.r.t. x

$$\frac{Q}{Q+1}$$