Question

evaluate i) ∫x²(1-x)²dx ii)∫x[sin²(sin x) + cos²(cos x)] dx

please give correct answer

Answer 1

ii) i was unable to solve 2nd one sorry......

Answer 2

Answer:

(i) $\tt{ \int {x^{2} (1-x)^{2}} \, dx}$ = $\tt{ \dfrac{x^{5}}{5} - \dfrac{x^{4}}{2} + \dfrac{x^{3}}{3} + c }$

(ii) $\tt{ \int\limits^\pi_0 {x [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$ = $\tt{ \dfrac{\pi^{2}}{2} }$

Step-by-step explanation:

(i) Given: $\tt{ \int {x^{2} (1-x)^{2}} \, dx}$

= $\tt{ \int {x^{2} (1+x^{2}-2x)} \, dx}$

= $\tt{ \int {(x^{2} +x^{4}-2x^{3})} \, dx}$

= $\tt{ \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - 2\dfrac{x^{4}}{4} + c }$

= $\tt{ \dfrac{x^{5}}{5} - \dfrac{x^{4}}{2} + \dfrac{x^{3}}{3} + c }$

(ii) Given: $\tt{ \int\limits^\pi_0 {x [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$

Let, I = $\tt{ \int\limits^\pi_0 {x [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$     --------[1]

⇒ I = $\tt{ \int\limits^\pi_0 {(\pi -x) [sin^{2}(sin(\pi-x)) + cos^{2}(cos(\pi-x))]} \, dx }$

⇒ I = $\tt{ \int\limits^\pi_0 {(\pi -x) [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$      --------[2]

On adding equations [1] and [2], we get

2I = $\tt{ \int\limits^\pi_0 {x [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx } + \tt{ \int\limits^\pi_0 {(\pi -x) [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$

⇒ 2I = $\tt{ \int\limits^\pi_0 {x [sin^{2}(sinx) + cos^{2}(cosx)] + (\pi -x) [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$

⇒ 2I = $\tt{ \int\limits^\pi_0 {(x+\pi-x) [sin^{2}(sinx) + cos^{2}(cosx)] } \, dx }$

⇒ 2I = $\tt{ \int\limits^\pi_0 { \pi [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$

⇒ 2I = $\tt{ 2\pi \int\limits^\frac{\pi}{2}_0 { [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$

⇒ I = $\tt{ \pi \int\limits^\frac{\pi}{2}_0 { [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$      --------[3]

⇒ I = $\tt{ \pi \int\limits^\frac{\pi}{2}_0 { [sin^{2}(sin(\dfrac{\pi}{2}-x)) + cos^{2}(cos(\dfrac{\pi}{2}-x))]} \, dx }$

⇒ I = $\tt{ \pi \int\limits^\frac{\pi}{2}_0 { [sin^{2}(cosx) + cos^{2}(sinx)]} \, dx }$      --------[4]

On adding equations [3] and [4], we get

2I = $\tt{ \pi \int\limits^\frac{\pi}{2}_0 { [sin^{2}(sinx) + cos^{2}(cosx)]} \, dx } + \tt{ \pi \int\limits^\frac{\pi}{2}_0 { [sin^{2}(cosx) + cos^{2}(sinx)]} \, dx }$

⇒ 2I = $\tt{ \pi \int\limits^\frac{\pi}{2}_0 { [sin^{2}(cosx) + cos^{2}(sinx) + sin^{2}(sinx) + cos^{2}(cosx)]} \, dx }$

⇒ 2I = $\tt{ \pi \int\limits^\frac{\pi}{2}_0 { 2} \, dx }$

⇒ I = $\tt{ \pi \int\limits^\frac{\pi}{2}_0 { 1} \, dx }$

⇒ I = $\tt{ \pi [\dfrac{\pi}{2} - 0 ] }$

⇒ I = $\tt{ \dfrac{\pi^{2}}{2} }$

Formulas used :-

1) sin²θ + cos²θ = 1

2) $\tt{ \int {x^{n}} \, dx = \dfrac{x^{n+1}}{n+1} }$

3) $\tt{ \int\limits^a_0 {f(x)} \, dx = \int\limits^a_0 {f(a-x)} \, dx }$

4) $\tt{ \int\limits^{2a}_{0} {f(x)} \, dx = 2 \int\limits^a_0 {f(x)} \, dx }$ , if f(x) is an even function.

5) $\tt{ \int\limits^{2a}_{0} {f(x)} \, dx = 0}$ , if f(x) is an odd function.

6) $\tt{ \int\limits^{a}_{0} {1} \, dx = a}$