Evaluate the integral: I = integral dx/(sin²(2x + pi/4))

Evaluate the integral: $$ I = \int \frac{dx}{\sin^2\left(2x + \frac{\pi}{4}\right)}. $$

Step 1 · Recognize the reciprocal identity

To solve the given integral, we start by recognizing that the integrand involves $\sin^2(u)$, where $u = 2x+\dfrac{\pi}{4}$. We can use the identity for the cosecant function:

$$ \csc^2(u) = \frac{1}{\sin^2(u)}. $$

Thus, we can rewrite the integrand as:

$$ \frac{1}{\sin^2\left(2x+\frac{\pi}{4}\right)} = \csc^2\left(2x+\frac{\pi}{4}\right). $$

Step 2 · Substitute

Next, let us make a substitution:

$$ u = 2x + \frac{\pi}{4}. $$

The derivative of $u$ with respect to $x$ is:

$$ \frac{du}{dx} = 2, $$

so that $dx = \dfrac{du}{2}$.

Step 3 · Transform and integrate

Substituting this into the integral, we have:

$$ I = \int \csc^2(u)\cdot \frac{du}{2} = \frac{1}{2}\int \csc^2(u)\,du. $$

The integral of $\csc^2(u)$ is a standard result:

$$ \int \csc^2(u)\,du = -\cot(u). $$

Therefore, the integral becomes:

$$ I = \frac{1}{2}\left(-\cot(u)\right) + C, $$

where $C$ is the constant of integration.

Step 4 · Substitute back

Finally, substituting back $u = 2x+\dfrac{\pi}{4}$, we get the solution:

$$ I = -\frac{1}{2}\cot\left(2x+\frac{\pi}{4}\right) + C. $$

Summary Table

Additional Notes

• The domain of $I$ excludes points where $\sin\left(2x+\tfrac\pi4\right)=0$, i.e. $2x+\tfrac\pi4=k\pi$ for integer $k$.
• The additive constant $\tfrac\pi4$ inside the argument shifts the graph but does not affect the coefficient $\tfrac12$ out front, which comes only from the coefficient of $x$.
• The result can be checked by differentiating $-\tfrac12\cot\left(2x+\tfrac\pi4\right)$ with the chain rule and confirming it returns $\dfrac{1}{\sin^2\left(2x+\tfrac\pi4\right)}$.