MCQ
The value of $\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\big(\text{x}^3+\text{x}\cos\text{x}+\tan^5\text{x}+1\big)\text{dx},$ is:
- A$0$
- B$2$
- ✓$\pi$
- D$1$
✓
Answer
Correct option: C.
$\pi$
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\big(\text{x}^3+\text{x}\cos\text{x}+\tan^5\text{x}+1\big)\text{dx}$
$=\Big[\frac{\text{x}^4}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}+\Big[\text{x}\sin\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin\text{x}\text{dx}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}(\sec^2\text{x}-1)\text{dx}+\Big[\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{\pi^4}{64}-\frac{\pi^4}{64}+\frac{\pi}{2}-\frac{\pi}{2}-\Big[-\cos\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\text{ dx}+\frac{\pi}{2}+\frac{\pi}{2}$
$=+0+\Big[\frac{\tan^4\text{x}}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\text{ dx}$
$=\pi-\Big[\frac{\tan^2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\Big[-\log\big(\cos\text{x}\big)\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\pi$
$=\Big[\frac{\text{x}^4}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}+\Big[\text{x}\sin\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin\text{x}\text{dx}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}(\sec^2\text{x}-1)\text{dx}+\Big[\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{\pi^4}{64}-\frac{\pi^4}{64}+\frac{\pi}{2}-\frac{\pi}{2}-\Big[-\cos\text{x}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}\\+\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan^3\text{x}\text{ dx}+\frac{\pi}{2}+\frac{\pi}{2}$
$=+0+\Big[\frac{\tan^4\text{x}}{4}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\sec^2\text{x}\text{ dx}-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\tan\text{x}\text{ dx}$
$=\pi-\Big[\frac{\tan^2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}-\Big[-\log\big(\cos\text{x}\big)\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\pi$
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