📝 Integrals of tan^n x sec^m x (39 MCQs)

📖 Explanation: The exponent of tangent ($m = 3$) is odd, so the standard procedure is to split off $\sec x \tan x$ and use the identity $\tan^2 x = \sec^2 x - 1$ with the substitution $u = \sec x$. This reduces the integral to a polynomial in $u$. Splitting off $\sec^2 x$ is appropriate when the exponent of secant is even.

📖 Explanation: When the exponent of secant is even, the standard method is to split off $\sec^2 x$ and let $u = \tan x$, since the derivative of $\tan x$ is $\sec^2 x$. This simplifies the integral to a polynomial in $u$.

📖 Explanation: Since the exponent of tangent is even and secant is odd, the standard procedure (third case in Table 7.3.2) is to rewrite $\tan^2 x$ in terms of $\sec x$, which reduces the problem to integrating $\sec^3 x$ and $\sec x$. Substituting $u = \tan x$ would require a $\sec^2 x$ factor which is not present, and substituting $u = \sec x$ would require a $\sec x \tan x$ factor.

📖 Explanation: The exponent of secant is even, so the appropriate substitution is $u = \tan x$. Then $\sec^4 x = \sec^2 x \cdot \sec^2 x = (\tan^2 x + 1) \cdot \sec^2 x$, and $du = \sec^2 x dx$. Thus the integral becomes $\int u^4 (u^2 + 1) du$. The student's expression $\int (u^2+1)^2 u^2 du$ corresponds to $u = \tan x$ and $\sec^4 x = (\tan^2 x + 1)^2$ before converting the remaining $\sec^2 x$ to $du$.

📖 Explanation: Starting with $\int \tan^3 x \sec^3 x \, dx$, split off $\sec x \tan x$ to get $\int \tan^2 x \sec^2 x (\sec x \tan x) \, dx$. Using $\tan^2 x = \sec^2 x - 1$ and $u = \sec x$, we get $\int (u^2 - 1) u^2 \, du$. The $\sec^2 x$ term comes from $\tan^2 x + 1$, but since $m$ is odd we use the identity $\tan^2 x = \sec^2 x - 1$.

📖 Explanation: The derivative of $\tan x$ is $\sec^2 x$, and $\sec^2 x$ is present in the integrand. Therefore, $du = \sec^2 x dx$, and the integral becomes $\int u^3 \, du$. This is a straightforward Easy of the substitution method.

📖 Explanation: Let $u = \tan x$, then $du = \sec^2 x dx$. The integral becomes $\int u \, du = \frac{1}{2} u^2 + C = \frac{1}{2} \tan^2 x + C$.

📖 Explanation: For $\int \tan^2 x \sec^3 x \, dx$, both the tangent and secant exponents are even and odd respectively. The standard method is to rewrite $\tan^2 x = \sec^2 x - 1$, yielding $\int \sec^5 x \, dx - \int \sec^3 x \, dx$, which requires the reduction formula for secant. In contrast, the other integrals can be solved with a simple substitution.

📖 Explanation: The correct reduction formula for $\int \sec^n x \, dx$ is $\frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x dx$. This can be derived using integration by parts with $u = \sec^{n-2} x$ and $dv = \sec^2 x dx$.

📖 Explanation: The correct reduction formula is $\int \tan^n x dx = \frac{\tan^{n-1} x}{n-1} - \int \tan^{n-2} x dx$. This is derived by rewriting $\tan^n x = \tan^{n-2} x (\sec^2 x - 1)$. The first term integrates to $\frac{\tan^{n-1} x}{n-1}$, leaving the negative of the integral of $\tan^{n-2} x$.

📖 Explanation: Using the reduction formula for tangent with $n=3$: $\int \tan^3 x dx = \frac{\tan^{2} x}{2} - \int \tan x dx = \frac{1}{2} \tan^2 x - (-\ln|\cos x|) + C = \frac{1}{2} \tan^2 x + \ln|\cos x| + C$. Since $-\ln|\cos x| = \ln|\sec x|$, the answer can also be written as $\frac{1}{2} \tan^2 x - \ln|\sec x| + C$.

📖 Explanation: Using the reduction formula for secant with $n=3$: $\int \sec^3 x dx = \frac{\sec^{1} x \tan x}{2} + \frac{1}{2} \int \sec x dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| + C$.

📖 Explanation: Let's verify: $\int \tan^3 x \sec x dx = \int \tan^2 x (\sec x \tan x) dx = \int (\sec^2 x - 1)(\sec x \tan x) dx$. Let $u = \sec x$, $du = \sec x \tan x dx$. Then the integral becomes $\int (u^2 - 1) du = \frac{1}{3} u^3 - u + C = \frac{1}{3} \sec^3 x - \sec x + C$. The student's answer is correct.

📖 Explanation: Using the reduction formula for $\tan^n x$: $\int \tan^4 x dx = \frac{\tan^{3} x}{3} - \int \tan^2 x dx$. Then $\int \tan^2 x dx = \tan x - x + C$. Combining: $\frac{1}{3} \tan^3 x - (\tan x - x) + C = \frac{1}{3} \tan^3 x - \tan x + x + C$.

📖 Explanation: When $m$ is odd, the standard procedure is to split off a factor of $\sec x \tan x$ and use the identity $\tan^2 x = \sec^2 x - 1$. The substitution $u = \sec x$ then reduces the integrand to a polynomial in $u$. This method works because $du = \sec x \tan x dx$, which is precisely the factor split off.

📖 Explanation: When $n$ is even, the standard procedure is to split off a factor of $\sec^2 x$ and use the identity $\sec^2 x = 1 + \tan^2 x$. The substitution $u = \tan x$ then reduces the integrand to a polynomial in $u$. This method works because $du = \sec^2 x dx$, which is precisely the factor split off.

📖 Explanation: For $\int \tan^4 x \sec^3 x \, dx$, neither $m$ is odd nor $n$ is even. The standard method (case 3) is to rewrite the integrand in terms of $\sec x$ using $\tan^2 x = \sec^2 x - 1$, resulting in $\int (\sec^2 x - 1)^2 \sec^3 x dx = \int (\sec^7 x - 2\sec^5 x + \sec^3 x) dx$, which requires the reduction formula for $\sec^n x$.

📖 Explanation: Let $u = \sec x$. Then $du = \sec x \tan x dx$. The integral becomes $\int \tan^2 x \sec^4 x (\sec x \tan x) dx = \int (u^2 - 1) u^4 du = \int (u^6 - u^4) du = \frac{1}{7} u^7 - \frac{1}{5} u^5 + C = \frac{1}{7} \sec^7 x - \frac{1}{5} \sec^5 x + C$. This is also equivalent to $\frac{1}{5} \sec^5 x - \frac{1}{7} \sec^7 x + C$ with a different constant if we factor out $-\sec^5 x$.

📖 Explanation: The derivative of $\tan^3 x$ is $3 \tan^2 x \sec^2 x$, so the derivative of $\frac{1}{3} \tan^3 x$ is $\tan^2 x \sec^2 x = \tan^2 x (1 + \tan^2 x) = \tan^2 x + \tan^4 x$. The derivative of $\tan x$ is $\sec^2 x = 1 + \tan^2 x$. Summing gives $\tan^2 x + \tan^4 x + 1 + \tan^2 x = \tan^4 x + 2\tan^2 x + 1$. But wait, $\frac{d}{dx}(\frac{1}{3}\tan^3 x + \tan x) = \tan^2 x \sec^2 x + \sec^2 x = \sec^2 x (\tan^2 x + 1) = \sec^4 x$. However, $\sec^4 x = (1+\tan^2 x)^2 = 1 + 2\tan^2 x + \tan^4 x$. But $\int \sec^4 x dx = \int (1+\tan^2 x)\sec^2 x dx = \tan x + \frac{1}{3}\tan^3 x + C$. The derivative of $\frac{1}{3}\tan^3 x + \tan x$ is $\tan^2 x \sec^2 x + \sec^2 x = \sec^2 x(\tan^2 x + 1) = \sec^4 x$. So the derivative is $\sec^4 x = (1+\tan^2 x)^2 = 1 + 2\tan^2 x + \tan^4 x$. The provided options are not correct; the correct derivative is $\sec^4 x$. But if the student made an error, they might think the derivative is $\tan^4 x$.

📖 Explanation: Converting $\tan^m x \sec^n x$ to sines and cosines results in $\int \frac{\sin^m x}{\cos^{m+n} x} dx$, which is often a more complicated integral to evaluate than the original. The methods based on splitting off $\sec^2 x$ or $\sec x \tan x$ and using substitutions are generally more efficient. While the conversion approach can work, it is rarely the most direct path.

📖 Explanation: The derivative of $\tan x$ is $\sec^2 x$. Therefore, the integral of $\sec^2 x$ is $\tan x + C$. This is a fundamental integration formula.

📖 Explanation: The derivative of $\ln|\sec x|$ is $\frac{1}{\sec x} \cdot \sec x \tan x = \tan x$. Therefore, the integral of $\tan x$ is $\ln|\sec x| + C$.

📖 Explanation: The derivative of $\ln|\sec x + \tan x|$ is $\frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \sec x$. Therefore, the integral of $\sec x$ is $\ln|\sec x + \tan x| + C$.

📖 Explanation: The integrand $\tan^3 x \sec^2 x$ contains $\sec^2 x dx$, which is exactly $d(\tan x)$. A common mistake is to not recognize this and instead try to rewrite using identities, overcomplicating the problem. The correct and simplest approach is the substitution $u = \tan x$, which yields $\int u^3 du = \frac{1}{4} \tan^4 x + C$.

📖 Explanation: Let $u = \tan x$. Then $du = \sec^2 x dx$. We also need to convert one $\sec^2 x$ to $1 + \tan^2 x$. So $\sec^4 x dx = \sec^2 x \sec^2 x dx = (1+\tan^2 x)\sec^2 x dx = (1+u^2) du$. The limits: when $x=0$, $u=0$; when $x=\pi/4$, $u=1$. So the integral becomes $\int_0^1 (1+u^2) du = [u + \frac{1}{3} u^3]_0^1 = 1 + \frac{1}{3} = \frac{4}{3}$.

📖 Explanation: Let $u = \tan x$, $du = \sec^2 x dx$. The limits become $u(-\pi/4) = -1$, $u(\pi/4) = 1$. The integral becomes $\int_{-1}^{1} u^3 du = [\frac{1}{4} u^4]_{-1}^{1} = \frac{1}{4}(1 - 1) = 0$. Alternatively, $\tan^3 x \sec^2 x$ is an odd function over the symmetric interval $[-\pi/4, \pi/4]$, so the integral is 0.

📖 Explanation: Since $n=4$ is even, split off $\sec^2 x$ and let $u = \tan x$. Then $\sec^4 x dx = \sec^2 x \sec^2 x dx = (1+\tan^2 x)\sec^2 x dx = (1+u^2) du$. The integral becomes $\int u^5 (1+u^2) du = \int (u^5 + u^7) du = \frac{1}{6} u^6 + \frac{1}{8} u^8 + C = \frac{1}{6} \tan^6 x + \frac{1}{8} \tan^8 x + C$.

📖 Explanation: Let $u = \tan x$. Then $du = \sec^2 x dx$. The limits: $x=0 \to u=0$, $x=\pi/4 \to u=1$. The integral becomes $\int_0^1 u^3 du = [\frac{1}{4} u^4]_0^1 = \frac{1}{4}$. So the area is $\frac{1}{4}$.

📖 Explanation: The substitution $u = \tan x$ requires the presence of $\sec^2 x dx$. Here, we have $\sec^3 x dx = \sec x \cdot \sec^2 x dx$. The extra $\sec x$ cannot be expressed as a function of $u = \tan x$ alone, so the substitution is not effective. The correct approach is to split off $\sec x \tan x$ and use $u = \sec x$.

📖 Explanation: Let $u = \tan x$, then $du = \sec^2 x dx$. The integral becomes $\int u^5 du = \frac{1}{6} u^6 + C = \frac{1}{6} \tan^6 x + C$. This is a straightforward Easy of the substitution method.

📖 Explanation: The integral of $\sec^2 x$ is $\tan x$. Therefore, $\int_{\pi/4}^{\pi/3} \sec^2 x dx = [\tan x]_{\pi/4}^{\pi/3} = \tan(\pi/3) - \tan(\pi/4) = \sqrt{3} - 1$.

📖 Explanation: The substitution $u = \sec x$ would require $du = \sec x \tan x dx$, but the integrand has $\sec^2 x$, not $\sec x \tan x$. The substitution $u = \tan x$ works perfectly because $du = \sec^2 x dx$. Rewriting in terms of $\sec x$ and then using the reduction formula would also work but is unnecessary. Substituting $u = \tan^2 x$ is also not efficient as $du = 2\tan x \sec^2 x dx$, leaving an extra $\tan x$ factor. So the invalid method is $u = \sec x$.

📖 Explanation: Using the reduction formula $\int \sec^n x dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x dx$. For $n=6$: $\int \sec^6 x dx = \frac{\sec^4 x \tan x}{5} + \frac{4}{5} \int \sec^4 x dx$. For $n=4$: $\int \sec^4 x dx = \frac{\sec^2 x \tan x}{3} + \frac{2}{3} \int \sec^2 x dx = \frac{\sec^2 x \tan x}{3} + \frac{2}{3} \tan x$. Substituting back: $\int \sec^6 x dx = \frac{1}{5} \sec^4 x \tan x + \frac{4}{5} [\frac{1}{3} \sec^2 x \tan x + \frac{2}{3} \tan x] + C = \frac{1}{5} \sec^4 x \tan x + \frac{4}{15} \sec^2 x \tan x + \frac{8}{15} \tan x + C$.