Integral Evaluation of Trigonometric Function Combinations

To determine the correct options, we start by simplifying the given function $f(x)$:

$f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x$

Factoring out common terms: $f(x) = 7\tan^6 x (\tan^2 x + 1) - 3\tan^2 x (\tan^2 x + 1)$ $f(x) = (7\tan^6 x - 3\tan^2 x)(\tan^2 x + 1)$

Using the identity $1 + \tan^2 x = \sec^2 x$, we get: $f(x) = (7\tan^6 x - 3\tan^2 x)\sec^2 x$

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Step 1: Evaluating $\int_{0}^{\pi/4} f(x) \, dx$

Let $I_1 = \int_{0}^{\pi/4} f(x) \, dx$: $I_1 = \int_{0}^{\pi/4} (7\tan^6 x - 3\tan^2 x)\sec^2 x \, dx$

Using the substitution $u = \tan x$, we have $du = \sec^2 x \, dx$. The limits of integration transform as follows:

• At $x = 0$, $u = 0$
• At $x = \frac{\pi}{4}$, $u = 1$

Substituting these into the integral gives: $I_1 = \int_{0}^{1} (7u^6 - 3u^2) \, du$ $I_1 = \left[ u^7 - u^3 \right]_{0}^{1} = (1^7 - 1^3) - 0 = 0$

Thus, Option (B) is correct and Option (D) is incorrect.

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Step 2: Evaluating $\int_{0}^{\pi/4} x f(x) \, dx$

Let $I_2 = \int_{0}^{\pi/4} x f(x) \, dx$.

We apply Integration by Parts: $\int u \, dv = u v - \int v \, du$

Let:

• $u = x \implies du = dx$
• $dv = f(x) \, dx \implies v = \int f(x) \, dx = \tan^7 x - \tan^3 x$

Applying the formula: $I_2 = \left[ x(\tan^7 x - \tan^3 x) \right]_{0}^{\pi/4} - \int_{0}^{\pi/4} (\tan^7 x - \tan^3 x) \, dx$

Evaluating the boundary term:

• At $x = \frac{\pi}{4}$: $\frac{\pi}{4} \left(1^7 - 1^3\right) = 0$
• At $x = 0$: $0 \cdot \left(0^7 - 0^3\right) = 0$

Thus, the boundary term vanishes, leaving: $I_2 = \int_{0}^{\pi/4} (\tan^3 x - \tan^7 x) \, dx$

We can rewrite the integrand as: $\tan^3 x - \tan^7 x = \tan^3 x (1 - \tan^4 x)$ $= \tan^3 x (1 - \tan^2 x)(1 + \tan^2 x)$ $= (\tan^3 x - \tan^5 x)\sec^2 x$

Substituting this back into $I_2$: $I_2 = \int_{0}^{\pi/4} (\tan^3 x - \tan^5 x)\sec^2 x \, dx$

Using the substitution $t = \tan x$, where $dt = \sec^2 x \, dx$, the limits $x \in [0, \pi/4]$ change to $t \in [0, 1]$: $I_2 = \int_{0}^{1} (t^3 - t^5) \, dt$ $I_2 = \left[ \frac{t^4}{4} - \frac{t^6}{6} \right]_{0}^{1} = \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$

Thus, Option (A) is correct and Option (C) is incorrect.

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Conclusion

The correct options are (A) and (B).