Definite Integration
Definite Integration
nta_pyq_2025_apr
Grade 12
Question:
Let for $f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x$, $I_1 = \displaystyle\int_0^{\pi/4}f(x)\,dx$ and $I_2 = \displaystyle\int_0^{\pi/4}xf(x)\,dx$. Then $7I_1+12I_2$ is equal to:
Step-by-Step Solution
Key Concept: Factor: $f(x) = (7\tan^6 x - 3\tan^2 x)\sec^2 x$. Then $I_1 = \int_0^{\pi/4}(7\tan^6 x-3\tan^2 x)\sec^2 x\,dx$ evaluates to $0$ by direct computation, and $I_2 = 1/12$ via integration by parts.
$f(x) = (7\tan^6 x - 3\tan^2 x)(1+\tan^2 x) = (7\tan^6 x-3\tan^2 x)\sec^2 x.$
$$I_1 = \left[\frac{7\tan^7 x}{7}-\tan^3 x\right]_0^{\pi/4} = 1-1 = 0.$$
For $I_2$: IBP with $u=x$, $dv = f(x)dx$:
$$I_2 = \left[x(\tan^7 x-\tan^3 x)\right]_0^{\pi/4} - \int_0^{\pi/4}(\tan^7 x-\tan^3 x)dx = 0 - \int_0^{\pi/4}\tan^3 x(\tan^4 x-1)dx.$$
$$= \int_0^{\pi/4}(\tan^3 x-\tan^5 x)\sec^2 x\,dx = \left[\frac{\tan^4 x}{4}-\frac{\tan^6 x}{6}\right]_0^{\pi/4} = \frac{1}{4}-\frac{1}{6} = \frac{1}{12}.$$
$$7I_1+12I_2 = 0 + 12\cdot\frac{1}{12} = 1.$$
Correct Answer: 2