📝 Products of tangents and secants integrals (39 MCQs)

📖 Explanation: The standard method for integrating $\int \tan^m x \sec^n x \, dx$ when $m$ is odd is to split off a factor of $\sec x \tan x$ and set $u = \sec x$. This works because the derivative of $\sec x$ is $\sec x \tan x$, and the remaining $\tan^2 x$ terms can be converted to $\sec^2 x - 1$. Option B correctly identifies this procedure.

📖 Explanation: When the power of secant $n$ is even, the strategy is to split off $\sec^2 x$ and use $u = \tan x$. Since $d(\tan x) = \sec^2 x dx$, this substitution directly simplifies the integral. For $\int \tan^4 x \sec^4 x \, dx$, writing $\sec^4 x = \sec^2 x \cdot \sec^2 x$ and converting one $\sec^2 x$ to $1 + \tan^2 x$ makes the integral a polynomial in $u = \tan x$.

📖 Explanation: This is a direct Easy of the substitution $u = \tan x$, since $du = \sec^2 x dx$. The integral becomes $\int u^2 du = \frac{u^3}{3} + C = \frac{\tan^3 x}{3} + C$. This is a fundamental result that forms the basis for more complex integrations involving tangent and secant products.

📖 Explanation: For $u = \sec x$, $du = \sec x \tan x dx$. The given integral has $\sec^2 x$ and $\tan^3 x$, but no factor of $\tan x$ to pair with the $du$. While you could rewrite $\tan^2 x = \sec^2 x - 1$, the remaining $\tan x$ cannot be expressed easily in terms of $\sec x$ without a factor, making the substitution inefficient. The correct choice is $u = \tan x$ because $du = \sec^2 x dx$, which is present.

📖 Explanation: When the power of tangent $m$ is even and the power of secant $n$ is odd, neither the $u = \tan x$ nor $u = \sec x$ substitution works directly. The strategy is to convert all $\tan^2 x$ terms to $\sec^2 x - 1$, reducing the integral to a sum of powers of $\sec x$. This then requires the reduction formula for $\int \sec^n x \, dx$. Option C is a classic example of this case.

📖 Explanation: With $u = \sec x$, $du = \sec x \tan x dx$. The integral $\int \tan^3 x \sec^3 x dx = \int \tan^2 x \sec^2 x (\sec x \tan x dx) = \int (u^2 - 1) u^2 du = \int (u^4 - u^2) du = \frac{u^5}{5} - \frac{u^3}{3} + C$. The student's answer suggests they integrated $u^4 - u^2$ correctly, but missed the fact that $\tan^3 x \sec^3 x = \tan^2 x \sec^2 x \cdot \sec x \tan x$. They likely forgot to convert $\tan^2 x$ to $\sec^2 x - 1$.

📖 Explanation: Since $m = 2$ is even and $n = 1$ is odd, we use the identity $\tan^2 x = \sec^2 x - 1$. This transforms the integrand to $(\sec^2 x - 1) \sec x = \sec^3 x - \sec x$. Thus, the integral becomes $\int \sec^3 x \, dx - \int \sec x \, dx$, which requires reduction formulas for $\sec^3 x$. Both A and B represent correct equivalent forms.

📖 Explanation: The strategy depends on which substitution yields a simpler form. For odd $m$, splitting off $\sec x \tan x$ allows $u = \sec x$, since the derivative of $\sec x$ is $\sec x \tan x$. For even $n$, splitting off $\sec^2 x$ allows $u = \tan x$, since the derivative of $\tan x$ is $\sec^2 x$. Both methods aim to convert the remaining part of the integrand into a polynomial in the new variable.

📖 Explanation: Since $n = 4$ is even, we split off $\sec^2 x$ and use $u = \tan x$. $\int \tan^3 x \sec^4 x dx = \int \tan^3 x (\sec^2 x) (\sec^2 x) dx = \int \tan^3 x (1 + \tan^2 x) \sec^2 x dx = \int u^3 (1 + u^2) du = \frac{u^4}{4} + \frac{u^6}{6} + C = \frac{\tan^4 x}{4} + \frac{\tan^6 x}{6} + C$.

📖 Explanation: The derivative of $\tan x$ is $\sec^2 x$. The given integral is $\int \tan^5 x \sec x dx$. There is no $\sec^2 x$ factor present, so $du = \sec^2 x dx$ cannot be directly substituted. While you could rewrite $\tan^2 x = \sec^2 x - 1$, the integral would become a product of $\sec x$ terms with no factor to produce $du$. The correct approach is to use $u = \sec x$ after splitting off $\sec x \tan x$.

📖 Explanation: The substitution $u = \sec x$ is most effective when $m$ is odd because we can split off a $\sec x \tan x$ factor for $du$. In $\int \tan^3 x \sec^3 x dx$, we have $m = 3$ (odd) and $n = 3$ (odd), so splitting off $\sec x \tan x$ is the standard strategy. For the other options, $n$ is even (Options A, C, D) which suggests $u = \tan x$ as a better choice.

📖 Explanation: When $n$ is even, the substitution $u = \tan x$ transforms the integral into a polynomial in $u$, which can be integrated directly. For $\int \tan^4 x \sec^4 x dx$, $n = 4$ is even, so $\sec^4 x = \sec^2 x \sec^2 x = (1+\tan^2 x) \sec^2 x$. With $u = \tan x$, this becomes a polynomial $\int u^4 (1+u^2) du$, requiring no reduction formula. The other options involve odd powers of $\sec x$ and thus require the reduction formula for $\sec^n x$.

📖 Explanation: The expression $\sec^2 x = 1 + \tan^2 x$, and with $u = \tan x$, $du = \sec^2 x dx$. The integrand $(\sec^2 x - 1)^2 \sec^2 x = (\tan^2 x)^2 \sec^2 x = u^4 \sec^2 x$. Therefore, $\int (\sec^2 x - 1)^2 \sec^2 x dx = \int u^4 du = \frac{u^5}{5} + C$. However, note that the question is about the transformed integral. The correct transformed integral is $\int u^4 du$. The provided option B is $\int (u^2 - 1)^2 u^2 du$, which corresponds to $\int \tan^4 x \sec^4 x dx$.

📖 Explanation: With $u = \tan x$, $du = \sec^2 x dx$. The integrand is $\tan^4 x \sec^2 x = u^4 du$. This is a perfect match, leading to $\int u^4 du$. The other options would arise if the $\sec^2 x$ factor was more complex, such as $\sec^4 x$ or if we had converted $\tan^2 x$ to $\sec^2 x - 1$, which is unnecessary here.

📖 Explanation: If the integral after substitution is $\int (\sec^2 x - 1)^2 \sec x dx$, then the original integrand was $(\sec^2 x - 1)^2 \sec x = (\tan^2 x)^2 \sec x = \tan^4 x \sec x$. The student applied the identity to convert $\tan^4 x$ to $(\sec^2 x - 1)^2$. This is a valid step for an integral like $\int \tan^4 x \sec x dx$, which requires the reduction formula.

📖 Explanation: This is a straightforward Easy of the substitution $u = \sec x$, since $du = \sec x \tan x dx$. The integral becomes $\int \sec^2 x (\sec x \tan x dx) = \int u^2 du = \frac{u^3}{3} + C = \frac{\sec^3 x}{3} + C$. Note that Option A is identical to Option C; the correct answer is A.

📖 Explanation: Using $u = \tan x$, $du = \sec^2 x dx$. The limits transform: when $x = 0$, $u = 0$; when $x = \pi/4$, $u = 1$. The integral becomes $\int_0^1 u^3 du = \frac{u^4}{4} \Big|_0^1 = \frac{1}{4}$. This demonstrates how the definite integral can be easily evaluated after the substitution, without needing to convert back to $x$.

📖 Explanation: To use the substitution $u = \sec x$, we need a factor of $\sec x \tan x$. We can rewrite $\tan^5 x \sec^3 x = \tan^4 x \sec^2 x (\sec x \tan x)$. Then, using $\tan^2 x = \sec^2 x - 1$, we get $(\sec^2 x - 1)^2 \sec^2 x (\sec x \tan x)$. Both forms are equivalent and prepare the integrand for the substitution, leading to a polynomial in $u = \sec x$.

📖 Explanation: With $u = \sec x$, $du = \sec x \tan x dx$. We rewrite $\tan^3 x \sec^7 x = \tan^2 x \sec^6 x (\sec x \tan x) = (\sec^2 x - 1) \sec^6 x (\sec x \tan x) = (u^2 - 1) u^6 du$. The integral becomes $\int (u^2 - 1) u^6 du = \int (u^8 - u^6) du$.

📖 Explanation: Since $n = 4$ is even, the standard strategy is to use $u = \tan x$. This is because we can split off $\sec^2 x$ for $du$, and convert the remaining $\sec^2 x$ to $1 + \tan^2 x$. This reduces the integral to a polynomial in $\tan x$. Using $u = \sec x$ would be inefficient because we would need to handle the even power of $\sec x$ without a natural $\sec x \tan x$ factor to pair with $du$.

📖 Explanation: If we set $u = \sec x$, then $du = \sec x \tan x dx$. The integral becomes $\int \tan^2 x \sec^2 x dx = \int \tan x (u) du$. Since $\tan x = \sqrt{u^2 - 1}$ (for $x>0$), this gives $\int u \sqrt{u^2 - 1} du$. While this is a valid integral, it's more complicated than necessary. Using $u = \tan x$ gives $\int u^2 du$, which is far simpler. The option C identifies the correct transformation.

📖 Explanation: When $m$ is odd, we can split off $\sec x \tan x$ for $u = \sec x$, and the remaining $\tan^2 x$ terms can be converted to $\sec^2 x - 1$. When $n$ is even, we can split off $\sec^2 x$ for $u = \tan x$, and the remaining $\sec^2 x$ can be converted to $1 + \tan^2 x$. Since both conditions are met, either substitution will work, though one may be slightly more efficient depending on the specific exponents.

📖 Explanation: Since $m = 3$ is odd, use $u = \sec x$, $du = \sec x \tan x dx$. $\int \tan^3 x \sec x dx = \int \tan^2 x (\sec x \tan x dx) = \int (\sec^2 x - 1) du = \int (u^2 - 1) du = \frac{u^3}{3} - u + C = \frac{\sec^3 x}{3} - \sec x + C$. This is a classic example of the $u = \sec x$ substitution.

📖 Explanation: Option B has $m = 5$ (odd) and $n = 5$ (odd). This means both substitutions are possible, but after using $u = \sec x$, we convert $\tan^4 x$ to $(\sec^2 x - 1)^2$, resulting in a polynomial in $\sec x$. This polynomial will include terms like $\sec^5 x$, $\sec^3 x$, and $\sec x$, requiring the reduction formula for $\sec^5 x$ and $\sec^3 x$. Options A, C, and D have even $n$, allowing for a simpler $u = \tan x$ substitution that yields a direct polynomial integral.

📖 Explanation: This is a direct Easy of the substitution $u = \sec x$, since $du = \sec x \tan x dx$. The integral becomes $\int \sec^4 x (\sec x \tan x dx) = \int u^4 du = \frac{u^5}{5} + C = \frac{\sec^5 x}{5} + C$. This is a fundamental result. Note that Option A is identical to Option C; the correct answer is A.

📖 Explanation: After converting $\tan^4 x$ to $(\sec^2 x - 1)^2$, the integrand becomes a sum of terms like $\sec^5 x$, $\sec^3 x$, and $\sec x$. The student must now integrate these powers of $\sec x$. This is done by using the reduction formula for $\int \sec^n x dx$, which reduces the power step by step until $\int \sec x dx = \ln|\sec x + \tan x| + C$ is reached.

📖 Explanation: This is a direct Easy of the substitution $u = \tan x$, since $du = \sec^2 x dx$. The integral becomes $\int u^2 du = \frac{u^3}{3} + C = \frac{\tan^3 x}{3} + C$. This is a fundamental result that forms the basis for more complex integrations involving tangent and secant products. Option A is identical to Option D; the correct answer is A.

📖 Explanation: With $u = \tan x$, $du = \sec^2 x dx$. The integral becomes $\int u^3 du = \frac{u^4}{4} + C = \frac{\tan^4 x}{4} + C$. Option C is identical to Option A. This is a direct substitution problem, similar to the previous one.

📖 Explanation: When $m$ is even and $n$ is odd, the standard strategy is to use the identity $\tan^2 x = \sec^2 x - 1$ to rewrite the integrand entirely in terms of $\sec x$. Then, the reduction formula for $\sec^n x$ is applied. The substitution $u = \tan x$ is NOT useful here because there is no $\sec^2 x$ factor to pair with $du$, and the resulting integral would involve odd powers of $\sec x$ that are difficult to express in terms of $\tan x$ alone. Option D incorrectly suggests using $u = \tan x$.

📖 Explanation: $\int \tan^2 x \sec x dx = \int (\sec^2 x - 1) \sec x dx = \int \sec^3 x dx - \int \sec x dx$. The reduction formula is needed for $\sec^3 x$. The integral of $\sec x$ is known. The reduction formula for $\sec^n x$ reduces the power by 2. So, starting from $\sec^3 x$, it will require the integral of $\sec x$. Therefore, both are needed.

📖 Explanation: With $u = \sec x$, $du = \sec x \tan x dx$. We rewrite $\tan^5 x \sec^4 x = \tan^4 x \sec^3 x (\sec x \tan x) = (\sec^2 x - 1)^2 \sec^3 x du = (u^2 - 1)^2 u^3 du$. The integral becomes $\int (u^2 - 1)^2 u^3 du = \int (u^7 - 2u^5 + u^3) du$. Option A is correct.

📖 Explanation: Since $m = 3$ is odd, the standard method is to split off $\sec x \tan x$ and use $u = \sec x$. This works because the derivative of $\sec x$ is $\sec x \tan x$. With $u = \sec x$, we have $\tan^2 x = u^2 - 1$, which simplifies the integrand to a polynomial in $u$. The integral becomes $\int (u^2 - 1) u^4 du = \int (u^6 - u^4) du$, which is easy to integrate.

📖 Explanation: Using $u = \sec x$, $du = \sec x \tan x dx$. The limits transform: when $x = 0$, $u = 1$; when $x = \pi/4$, $u = \sqrt{2}$. The integral becomes $\int_1^{\sqrt{2}} (u^2 - 1) du = \frac{u^3}{3} - u \Big|_1^{\sqrt{2}} = (\frac{2\sqrt{2}}{3} - \sqrt{2}) - (\frac{1}{3} - 1) = \frac{2\sqrt{2}}{3} - \sqrt{2} - \frac{1}{3} + 1 = \frac{2\sqrt{2}}{3} - \frac{3\sqrt{2}}{3} + \frac{2}{3} = \frac{2\sqrt{2} - 3\sqrt{2} + 2}{3} = \frac{2 - \sqrt{2}}{3}$. Wait, this doesn't match Option A. Let's recalculate: $\frac{u^3}{3} - u$ from 1 to $\sqrt{2}$ is $(\frac{2\sqrt{2}}{3} - \sqrt{2}) - (\frac{1}{3} - 1) = \frac{2\sqrt{2}}{3} - \frac{3\sqrt{2}}{3} - \frac{1}{3} + 1 = \frac{-\sqrt{2}}{3} + \frac{2}{3} = \frac{2 - \sqrt{2}}{3}$. Option A is $\frac{2\sqrt{2}}{3} - 1$. There might be a mistake. Let's check the integral: $\int \tan^3 x \sec x dx = \int \tan^2 x \sec x \tan x dx = \int (\sec^2 x - 1) \sec x \tan x dx$. With $u = \sec x$, $du = \sec x \tan x dx$, this is $\int (u^2 - 1) du = \frac{u^3}{3} - u$. Evaluating from 1 to $\sqrt{2}$: $(\frac{2\sqrt{2}}{3} - \sqrt{2}) - (\frac{1}{3} - 1) = \frac{2\sqrt{2}}{3} - \frac{3\sqrt{2}}{3} - \frac{1}{3} + 1 = -\frac{\sqrt{2}}{3} + \frac{2}{3} = \frac{2 - \sqrt{2}}{3}$. So the correct answer is $\frac{2 - \sqrt{2}}{3}$. Option A is incorrect.