Question

\(\int\frac{sin^{25}x}{cos^{27}x}dx\) is equal to

• A. \(\frac{\sin^{26}(x)}{26}+C\)
• B. \(\frac{\cos^{26}(x)}{26}+C\)
• C. \(\tan^{26}(x)+C\)
• D. \(\frac{\tan^{26}(x)}{26}+C\)
• E. \(26\tan^{26}(x)+C\)

Answer 1

The Correct Option is D

We are given the integral: \[ \int \frac{\sin^{25} x}{\cos^{27} x} \, dx \] We can express the integrand as: \[ \frac{\sin^{25} x}{\cos^{27} x} = \frac{\sin^{25} x}{\cos^{26} x} \cdot \frac{1}{\cos x} \] Now, let's separate one factor of $\sin x$ and use the identity: \[ \sin x = \frac{d}{dx} (-\cos x) \] Thus, we rewrite the integral as: \[ \int \frac{\sin^{24} x}{\cos^{26} x} \cdot \sin x \, dx \] Substituting the identity, we get: \[ \int \tan^{26} x \, dx \] The antiderivative of $\tan^{26} x$ is: \[ \frac{\tan^{26} x}{26} + C \]

The correct option is (D) : \(\frac{\tan^{26}(x)}{26}+C\)

Answer 2

We want to evaluate the integral: \[\int \frac{\sin^{25}x}{\cos^{27}x} dx\]

We can rewrite the integrand as: \[\frac{\sin^{25}x}{\cos^{27}x} = \frac{\sin^{25}x}{\cos^{25}x \cdot \cos^2 x} = \frac{\sin^{25}x}{\cos^{25}x} \cdot \frac{1}{\cos^2 x} = \tan^{25}x \cdot \sec^2 x\]

So the integral becomes: \[\int \tan^{25}x \sec^2 x \, dx\]

Now, we use the substitution method. Let \(u = \tan x\). Then, the derivative of \(u\) with respect to \(x\) is: \[\frac{du}{dx} = \sec^2 x\] \[du = \sec^2 x \, dx\]

Substitute \(u\) and \(du\) into the integral: \[\int \tan^{25}x \sec^2 x \, dx = \int u^{25} \, du\]

Integrate with respect to \(u\): \[\int u^{25} \, du = \frac{u^{26}}{26} + C\]

Finally, substitute back \(u = \tan x\): \[\frac{\tan^{26}x}{26} + C\]

Therefore, the integral is: \[\int \frac{\sin^{25}x}{\cos^{27}x} dx = \frac{\tan^{26}x}{26} + C\]