Trigonometric Identities and Trigonometric Substitution

Practice flashcards in fill-in-the-blank format covering essential trigonometric identities, derivatives, and trigonometric substitution cases with worked examples.

According to the Pythagorean identities, $\sin^2(x) + \cos^2(x) =$ __________.

$$1$$

The Pythagorean identity relating secant and tangent is $\sec^2(x) =$ __________.

$$1 + \tan^2(x)$$

The Pythagorean identity relating cosecant and cotangent is $\csc^2(x) =$ __________.

$$1 + \cot^2(x)$$

The double-angle formula for sine states that $\sin(2x) =$ __________.

$$2\sin(x)\cos(x)$$

The double-angle formula for cosine states that $\cos(2x) =$ __________.

$$\cos^2(x) - \sin^2(x)$$

The power-reducing formula for sine squared is $\sin^2(x) =$ __________.

$$\frac{1 - \cos(2x)}{2}$$

The product-to-sum identity for $\sin(A)\sin(B)$ is __________.

$$\frac{1}{2}(\cos(A - B) - \cos(A + B))$$

The product-to-sum identity for $\cos(A)\cos(B)$ is __________.

$$\frac{1}{2}(\cos(A - B) + \cos(A + B))$$

The derivative of $\tan(x)$ is __________.

$$\sec^2(x)$$

The derivative of $\cot(x)$ is __________.

$$-\csc^2(x)$$

The derivative of $\sec(x)$ is __________.

$$\sec(x)\tan(x)$$

The derivative of $\csc(x)$ is __________.

$$-\csc(x)\cot(x)$$

When simplifying an integrand with the expression $\sqrt{1 - x^2}$, the standard trigonometric substitution is $x =$ __________.

$$\sin(\theta)$$

When using the trigonometric substitution $x = \sin(\theta)$, the radical expression $\sqrt{1 - \sin^2(\theta)}$ simplifies to __________.

$$\cos(\theta)$$

For an integrand involving $\sqrt{1 + x^2}$, the recommended trigonometric substitution is $x =$ __________.

$$\tan(\theta)$$

For an integrand involving $\sqrt{x^2 - 1}$, the recommended trigonometric substitution is $x =$ __________.

$$\sec(\theta)$$

Evaluating $\int \frac{1}{\sqrt{1 - x^2}}\,dx$ by substituting $x = \sin(\theta)$ produces the result __________.

$$\sin^{-1}(x) + C$$

To evaluate $\int \frac{1}{1 + 16x^2}\,dx$ using Case 2 substitution, setting $16x^2 = \tan^2(\theta)$ yields the substitution $x =$ __________.

$$\frac{1}{4}\tan(\theta)$$

The evaluation of $\int \frac{1}{1 + 16x^2}\,dx$ results in __________.

$$\frac{1}{4}\tan^{-1}(4x) + C$$

To evaluate $\int \frac{1}{x\sqrt{4x^2 - 1}}\,dx$, setting $(2x)^2 - 1$ leads to the substitution $2x =$ __________.

$$\sec(\theta)$$

The evaluation of $\int \frac{1}{x\sqrt{4x^2 - 1}}\,dx$ produces the final expression __________.

$$\sec^{-1}(2x) + C$$

Using the Right Triangle Trick for $\sin(\theta) = \frac{2x}{1}$, the value of $\cot(\theta)$ is expressed as __________.

$$\frac{\sqrt{1 - 4x^2}}{2x}$$

When handling $\int \sqrt{x^2 - 25}\,dx$, multiplying and dividing inside the radical by $25$ simplifies the expression to __________.

$$5\sqrt{\frac{x^2}{25} - 1}$$

For the integral $5\int \sqrt{\frac{x^2}{25} - 1}\,dx$, setting $\frac{x}{5} = \sec(\theta)$ gives $dx =$ __________.

$$5\sec(\theta)\tan(\theta)\,d\theta$$

The antiderivative of $\sec(\theta)$ with respect to $\theta$ is __________.

$$\ln|\sec(\theta) + \tan(\theta)| + C$$